284185717-Munkres-Solucionario.pdf

Selected Solutions to Munkres’s Topology, 2nd Ed. Takumi Murayama December 20, 2014 These solutions are the result of taking MAT365 Topology in the Fall of 2012 at Princeton University. This is not a complete set of solutions; see the List of Solved Exercises at the end. Please e-mail [email protected] with any corrections.

Contents I

General Topology

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1 Set Theory and Logic 7 Countable and Uncountable Sets . . . . . . . . . . . . . . . . . . . .

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2 Topological Spaces and Continuous Functions 13 Basis for a Topology . . . . . . . . . . . . . . . 16 The Subspace Topology . . . . . . . . . . . . . 17 Closed Sets and Limit Points . . . . . . . . . . 18 Continuous Functions . . . . . . . . . . . . . . 19 The Product Topology . . . . . . . . . . . . . 20 The Metric Topology . . . . . . . . . . . . . . 21 The Metric Topology (continued) . . . . . . . . 22 The Quotient Topology . . . . . . . . . . . . .

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3 Connectedness and Compactness 23 Connected Spaces . . . . . . . . . . . 24 Connected Subspaces of the Real Line 25 Components and Local Connectedness 27 Compact Subspaces of the Real Line . 29 Local Compactness . . . . . . . . . .

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4 Countability and Separation Axioms 30 The Countability Axioms . . . . . . 31 The Separation Axioms . . . . . . . 32 Normal Spaces . . . . . . . . . . . . 33 The Urysohn Lemma . . . . . . . . 34 The Urysohn Metrization Theorem . 36 Imbeddings of Manifolds . . . . . .

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Algebraic Topology

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9 The 51 52 53 54 58 59 60

Fundamental Group Homotopy of Paths . . . . . . . . . . . . . The Fundamental Group . . . . . . . . . . Covering Spaces . . . . . . . . . . . . . . . The Fundamental Group of the Circle . . . Deformation Retracts and Homotopy Type The Fundamental Group of S n . . . . . . . Fundamental Groups of Some Surfaces . . .

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Seifert-van Kampen Theorem Direct Sums of Abelian Groups . . . . . . . . . . . . . . . . Free Products of Groups . . . . . . . . . . . . . . . . . . . The Fundamental Group of a Wedge of Circles . . . . . . . The Fundamental Groups of the Torus and the Dunce Cap

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12 Classification of Surfaces 49 74 Fundamental Groups of Surfaces . . . . . . . . . . . . . . . . . . . . 49 75 Homology of Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . 49

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Part I

General Topology 1 7

Set Theory and Logic Countable and Uncountable Sets

Exercise 7.5. Determine, for each of the following sets, whether or not it is countable. Justify your answers. (j) The set J of all finite subsets of Z+ . S Solution for (j). We claim J is countable. Consider I = ∞ n=0 In where In is the set of sequences with n elements. Each In is countable by Theorem 7.6 so I is countable by Theorem 7.5. Identifying each finite subset in J with the finite sequence with the same elements in increasing order, we see that J ⊆ I, and so J is countable by Corollary 7.3.

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Topological Spaces and Continuous Functions

13

Basis for a Topology

Exercise 13.3. Show that the collection Tc given in Example 4 of §12 is a topology on the set X. Is the collection T∞ = {U | X \ U is infinite or empty or all of X} a topology on X? Proof. Recall Example 12.4: Let X be a set; let Tc be the collection of all subsets U of X such that X \ U either is countable or is all of X. We claim this forms a topology on X; we will follow the numbering for the definition of a topology on p. 76. S T (1) X \ ∅ = X and X \ X = ∅ is countable; (2) X \ α∈A Uα = α∈A (X \ Uα ) is countable S since it is an intersectionTof countable sets, S unless every Uα = ∅, in which case X \ α∈A Uα = X; (3) X \ α∈A finite Uα = α∈A finite (X \ Uα ) is countable sinceTit is the finite union of countable sets, unless every Uα = ∅, in which case X \ α∈A finite Uα = X. Now consider T∞ . It is not a topology, for if we let X = [−1, 1] ⊆ R, and let U1 = [−1, 0) and U2 = (0, 1], we see that both U1 , U2 ∈ T∞ , but X \ (U1 ∪ U2 ) = {0}, which is not infinite, and so U1 ∪ U2 ∈ / T∞ . 3

Exercise 13.5. Show that if A is a basis for a topology on X, then the topology generated by A equals the intersection of all topologies on X that contain A. Prove the same if A is a subbasis. Proof. Let TA be the topology generated by A, and TI be the intersection of all topologies that contain A. TI ⊆ TA . This follows from the fact that TA ⊇ A, and so is one of the topologies that is intersected over in the construction of TI S . TA ⊆ T I . S Let U ∈ TA ; by Lemma 13.1, U = α Aα for some collection {Aα }α ⊆ A. But U = α Aα ∈ TI since each Aα ∈ TI . Now let A be a subbasis. The proof that TI ⊆ TA is identical; it remains to show TA ⊆ TI . Let U ∈ TA ; by definition of the topology generated by A, U is the union of a finite intersection of elements {Aα }α ⊆ A. But then U ∈ TI since each Aα ∈ TI . Exercise 13.6. Show that the topologies of R` and RK are not comparable. Proof. R` 6⊆ RK . For [a, b) ∈ R` , there is no basis element U ∈ RK such that a ∈ U, U ⊆ [a, b). RK 6⊆ R` . For (−1, 1) \ K ∈ RK which contains 0, there is no basis element [a, b) ∈ R` such that 0 ∈ [a, b), [a, b) ⊆ (−1, 1) \ K by the Archimedean property, that is, for all > 0, there exists N ∈ N such that 1/N < . Exercise 13.7. Consider the following topologies on R: T1 T2 T3 T4 T5

= the = the = the = the = the

standard topology, topology of RK , finite complement topology, upper limit topology, having all sets (a, b] as basis, topology having all sets (−∞, a) = {x | x < a} as basis.

Determine, for each of these topologies, which of the others it contains. Proof. We claim we have the following Hasse diagram: T4 T2 T1 T3

T5 4

T3 ( T1 . Inclusion is true since U ∈ST3 =⇒ U c finite, and so if we let U c = {xi }ni=1 with xi in increasing order, U = ni=0 (xi , xi+1 ), where x0 = −∞, xn+1 = ∞. Inequality follows since for (a, b) such that −∞ < a, b < ∞, R \ (a, b) is not finite. T5 ( T1 . Inclusion is clear since (−∞, a) is of the form (b, c). Inequality follows since for (b, c) ∈ T1 and x ∈ (b, c), there is no basis element (−∞, a) ∈ T5 such that x ∈ (−∞, a), (−∞, a) ⊆ (b, c) (if b > −∞). T3 and T5 are not comparable. T3 6⊆ T5 since R \ {0} ∈ T3 , but if we take x > 0, which is in this set, there is no basis element (−∞, a) ∈ T5 that contains x but is contained in R \ {0}. T5 6⊆ T3 since (−∞, 0)c is not finite. T1 ( T2 by Lemma 13.4. T2 ( T4 . For (a, b) ∈ T2 and x ∈ (a, b), (a, x] ∈ T4 and (a, x] ⊆ (a, b). For (a, b) \ K ∈ T2 and x ∈ (a, b) \ K, we note that x ∈ (1/(n + 1), c] where x < c < 1/n, x ∈ (a, 0], or x ∈ (1, d], where x < d < b; in all three cases, these sets are subsets of (a, b) \ K and are members of T4 . Inequality follows since for (a, b] ∈ T4 , there is no basis element U ∈ T2 such that b ∈ U, U ⊂ (a, b].

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The Subspace Topology

Exercise 16.8. If L is a straight line in the plane, describe the topology L inherits as a subspace of R` × R and as a subspace of R` × R` . In each case it is a familiar topology. Solution. Note that the basis for R` ×R consists of elements of the form [a, b)×(c, d). If L = {(x, y) | x = x0 }, then L∩[a, b)×(c, d) = ∅ or {x0 }×(c, d), and so defining the map ϕ : L ∩ (R` × R) → R, {x0 } × (c, d) 7→ (c, d), it is bijective, open, and continuous, and so the topology L inherits is homeomorphic to R with the standard topology. If L has finite slope, we first note that L ∩ (R` × R) = {(x, mx + b) ∈ R2 | x ∈ R}, and that the basis for our topology are the sets of the form ∅, [(a, ma + b), (c, mc + b)), ((a, ma + b), (c, mc + b)) for a, c ∈ R and a < c, by Lemma 16.1. We then define ϕ : L ∩ (R` × R) → R` ,

(a, ma + b) 7→ a.

This implies ((a, ma + b), (c, mc + b)) 7→ (a, c), [(a, ma + b), (c, mc + b)) 7→ [a, c). We claim this defines a homeomorphism with R` . Clearly, it is continuous, for the basis elements of R` have preimages that are basis elements in the topology on L. Likewise, it is open since the basis elements of L map to sets that are open in R` by 5

Lemma 13.4. Finally this is a bijection since there exists an inverse just by reversing the arrows above. For R` × R` , by following the same steps as above if L = {(x, y) | x = x0 }, then L ∩ (R` × R` ) is homeomorphic to R` . For L with |m| < ∞, we must split it up into two cases. When m ≥ 0, we have a similar situation as above, except we only have to consider basis elements of the form [a, b); thus, L ∩ (R` × R` ) is homeomorphic to R` . When m < 0, since for every point (x, y) ∈ L, we can find a basis element [x, a) × [y, b) ∈ (R` × R` ) such that L ∩ [x, a) × [y, b) = {(x, y)}, and these form the open sets of our new topology by Lemma 16.1. We see then that the topology on L is homeomorphic to the discrete topology on R. Exercise 16.9. Show that the dictionary order topology on the set R × R is the same as the product topology Rd ×R, where Rd denotes R in the discrete topology. Compare this topology with the standard topology on R2 . Proof. We see that the basis elements of (R × R)lex consist of intervals of the form (a × b, c × d) for a < c, and for a = c and b < d, as in Example 14.2. These basis elements are open in Rd × R since (a × b, c × d) = (a, c) × R ∪ {a} × (c, ∞) ∪ {b} × (−∞, d) ∈ TRd ×R . For the reverse situation, consider the basis elements for Rd × R; these consist of all {a} × (b, c) since {a | a ∈ R} forms a basis for Rd by Example 13.3. But then, {a}×(b, c) are open in R×R with the order topology since it is of the form (a×b, c×d) for a = c. We now compare this to the standard topology on R2 . Since (a, b)×(c, d) ∈ Rd ×R, we see that R2 ⊆ Rd × R. Moreover, since {a} × (b, c) ∈ (Rd × R) \ R2 , we see that R2 ( Rd × R.

17

Closed Sets and Limit Points

Exercise 17.2. Show that if A is closed in Y and Y is closed in X, then A is closed in X. Proof. A is closed in Y iff there exists B ⊆ X closed in X such that A = Y ∩ B by Theorem 17.2. But then, A is the intersection of closed sets, and so is closed. Exercise 17.3. Show that if A is closed in X and B is closed in Y , then A × B is closed in X × Y .

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Proof. We see that X \A, Y \B are open in X, Y respectively by definition of a closed set. By definition of the product topology, (X \A)×Y, X ×(Y \B) are open in X ×Y . We see that (X \ A) × Y = (X × Y ) \ (A × Y ), X × (Y \ B) = (X × Y ) \ (X × B), and so A × Y, X × B are closed in X × Y . Finally, A × B = (A × Y ) ∩ (X × B), and so is the intersection of closed sets, i.e., closed. Exercise 17.5. Let X be an ordered set in the order topology. Show that (a, b) ⊂ [a, b]. Under what conditions does equality hold? Proof. Since (a, b) ⊆ [a, b] closed, and by the definition of closure, \ (a, b) = K ⊆ [a, b]. K⊇(a,b) closed

(a, b) = [a, b] ⇐⇒ a, b ∈ (a, b) ⇐⇒ any basis elements A 3 a, B 3 b intersect (a, b) by Theorem 17.5(b). We claim that this is equivalent to the fact that there is no immediate successor α of a and no immediate predecessor β of b. If either are the case, say for a, then choosing A with upper bound α would not intersect (a, b), and so equality doesn’t hold since a ∈ / (a, b); in the other direction, if neither are the case, we see that, say for a, the upper bound of A, α would be such that (a, α) is non-empty, and so A ∩ (a, b) 6= ∅, satisfying the condition for Theorem 17.5(b). The same argument applies when considering b and β, and so our claim holds. Exercise 17.13. Show that X is Hausdorff if and only if the diagonal ∆ = {x × x | x ∈ X} is closed in X × X. Proof. Suppose ∆ is closed in X × X, i.e., the complement ∆c is open. This is equivalent to for all (x, y) ∈ X × X such that x 6= y, there exists a basis element U × V of X × X for U, V open in X such that (x, y) ∈ U × V but (U × V ) ∩ ∆ = ∅. But then, by definition of ∆, this is equivalent to saying for all x, y ∈ X such that x 6= y, there exist open neighborhoods U 3 x and V 3 y such that U ∩ V = ∅, and so X is Hausdorff. Exercise 17.16. Consider the five topologies on R given in Exercise 7 of §13. (a) Determine the closure of the set K = {1/n | n ∈ Z+ } under each of these topologies. (b) Which of these topologies satisfy the Hausdorff axiom? the T1 axiom? Solution for (a). For T3 , A closed ⇐⇒ A finite or all of R. Since no finite set contains all of K, we see that R is the only closed set containing K, and so K = R. 7

For T5 , we claim K = [0, ∞). For x ∈ [0, ∞), the basis elements that contain x are of the form (−∞, a) for a > x. Since (−∞, a) ∩ K 6= ∅ by the Archimedean property, that is, ∀ > 0∃N ∈ N such that 1/N < , K = [0, ∞) by Theorem 17.5. For T1 , K 0 = {0} by Example 17.8, and so K = K ∪ {0}. For T2 , K is closed since R \ K = (−∞, ∞) \ K is a basis element, and so K = K. For T4 , K = K since T4 is finer than T2 , and so R \ K is still open. Solution for (b). T3 is T1 since any finite point set is closed by definition of the finite complement topology. It is not Hausdorff, for if we choose U 3 x, V 3 y both open, (U ∩ V )c = U c ∪ V c is finite, where the equality follows from De Morgan’s Laws, and so U ∩ V is infinite. T5 is not Hausdorff and not even T1 , for R \ {x0 } is not a union of basis elements, and so {x0 } is not closed. T1 is Hausdorff, for if we have x, y ∈ R and 0 < < |x − y|/2, then (x − , x + ) ∩ (y − , y + ) = ∅. Since Hausdorff =⇒ T1 , we see that T1 is also T1 . Since T2 , T4 are both finer than T1 , we see that the open sets constructed above are still open and separate x, y, and so T2 , T4 are still both Hausdorff and thus T1 .

18

Continuous Functions

Exercise 18.1. Prove that for functions f : R → R, the -δ definition of continuity implies the open set definition. Remark. Recall that f is continuous if for every > 0 and x0 ∈ R, there exists a δ > 0 such that |f (x) − f (x0 )| < for all x ∈ R such that |x − x0 | < δ. Proof. Consider x0 ∈ R, and a corresponding neighborhood V of f (x0 ); we then have V ⊇ (f (x0 ) − , f (x0 ) + ) for some > 0 since V is open. Then, by hypothesis there exists a δ > 0 such that f (x) ∈ (f (x0 ) − , f (x0 ) + ) for all x ∈ R such that x ∈ (x0 − δ, x0 + δ) = U , which is open. Thus, f (U ) ⊆ V , and so f is continuous by Theorem 18.1. Exercise 18.12. Let F : R × R → R be defined by the equation ( xy/(x2 + y 2 ) if x × y 6= 0 × 0. F (x × y) = 0 if x × y = 0 × 0. (a) Show that F is continuous in each variable separately. (b) Compute the function g : R → R defined by g(x) = F (x × x). (c) Show that F is not continuous. 8

Proof of (a). Since F is symmetric in interchanging x ↔ y, we only have to prove ∀y0 ∈ Y , h(x) = F (x × y0 ) is continuous as a function R → R. For y0 = 0, this is trivially true for the image of h is (0, 0) with preimage R. Now suppose y0 6= 0; then we have h(x) = xy0 /(x2 + y02 ). This is continuous since xy0 and x2 + y02 are both continuous, and so their quotient is also continuous (since also x2 + y02 6= 0), using the -δ definition of continuity (see Theorem 21.5). Proof of (b). Since F (x × x) for x 6= 0 equals x2 /(x2 + x2 ) = x2 /2x2 = 1/2, we have ( 1/2 if x 6= 0. g(x) = 0 if x = 0. Proof of (c). We claim F (x×y) is not continuous along the line L = {x = y} at (0, 0), i.e., F |L is not continuous at (0, 0); this suffices by Theorem 18.2(d). Note that the line L in the subspace topology is homeomorphic to R, where the homeomorphism is given by either of the coordinate projection maps. Now the preimage of the closed set {1/2} ⊆ R is L \ {(0, 0)}, which is not closed since R \ {0} is not closed, hence F |L is not continuous, and neither is F .

19

The Product Topology

Q Exercise 19.6. Let x1 , x2 , . . . be a sequence of the points of the product space Xα . Show that this sequence converges to the point x if and only if the sequence πα (x1 ), πα (x2 ), . . . converges to πα (x) for each α. Is this fact true if one uses the box topology instead of the product topology? Proof. Suppose {xiQ } → x, and fix some index γ. Then, for any neighborhood Uγ 3 πγ (x), letting U = Uα where Uα = Xα for all α 6= γ, there exists N ∈ N such that xi ∈ U for all i ≥ N , and so πγ (xi ) ∈ πγ (U ) = Uγ for all i ≥ N , i.e., {πγ (xi )} → πγ (x). Note that this direction does not depend on the topology being the product or box topology. In the other direction, Qsuppose {πα (xi )} → πα (x) for all α. We take Q an arbitrary neighborhood V of x ∈ Xα ; it then Q contains a basis element of Xα containing x, which is a product of open sets Uα . In the case of the product topology, there then exist only finite Uα ( Xα , and for these open sets there exist Nα ∈ N such that πα (xi ) ∈ Uα for all i ≥ Nα for each Q α. Nα = 1 works for all other α. Thus, we can take N = max(Nα ); then, xi ∈ Uα ⊆ V for all i ≥ N . We construct a counterexample for this direction in the case of the box topology. Let RN be the box topology on the product of copies of R indexed by N, and let xi := ( 1i , 1i , 1i , . . .). 9

Then, for each α ∈ N, {πα (xi )} → (0, 0, 0, . . .) =: x, but this sequence does not converge in the box topology, for the open set Y (− 1i , 1i ) = (−1, 1) × (− 12 , 12 ) × (− 13 , 13 ) × · · · i∈N

in the box topology contains x = (0, 0, 0, . . .), but does not contain any xi .

20

The Metric Topology

Exercise 20.4. Consider the product, uniform, and box topologies on Rω . (a) In which topologies are the following functions from R to Rω continuous? f (t) = (t, 2t, 3t, . . .), g(t) = (t, t, t, . . .), h(t) = (t, 12 t, 31 t, . . .). (b) In which topologies do the following sequences converge? w1 = (1, 1, 1, 1, . . .), w2 = (0, 2, 2, 2, . . .), w3 = (0, 0, 3, 3, . . .), ... y1 = (1, 0, 0, 0, . . .), y2 = ( 12 , 12 , 0, 0, . . .), y3 = ( 13 , 13 , 31 , 0, . . .), ...

x1 = (1, 1, 1, 1, . . .), x2 = (0, 12 , 21 , 12 , . . .), x3 = (0, 0, 31 , 13 , . . .), ... z1 = (1, 1, 0, 0, . . .), z2 = ( 21 , 12 , 0, 0, . . .), z3 = ( 31 , 13 , 0, 0, . . .), ...

Solution for (a). For the product topology, by Theorem 19.6, f, g, h are all continuous since each coordinate function is continuous. This is because if an open set in the image of a coordinate function is (a, b), its preimage would still be in the form (a0 , b0 ) ⊆ R where a0 , b0 are determined by the linear equations defining f, g, h above. Now consider the uniform topology. Note by Theorem 21.1 we can use the familiar -δ definition for continuity since our spaces both are metric spaces. We claim f is not continuous. For, suppose it is continuous. Then, given > 0 and x ∈ R, there exists δ > 0 such that |x−y| < δ =⇒ |f (x)−f (y)| = supn [min(n|x−y|, 1)] < . But, this is a contradiction since for n large, min(n|x−y|, 1) = 1, and so is always greater than . Now consider g. g is continuous since given > 0 and x ∈ R, we let δ < min(, 1) and 10

therefore have |x−y| < δ =⇒ |f (x)−f (y)| = supn [min(|x−y|, 1)] = min(|x−y|, 1) < min(, 1) ≤ . h is also continuous since given > 0 and x ∈ R, we let δ < min(, 1) and therefore have |x − y| < δ =⇒ |f (x) − f (y)| = supn [min(|x − y|/n, 1)] ≤ min(|x − y|, 1) < min(, 1) ≤ . For the box topology, since the box topology is finer than the uniform topology by Theorem 20.4, we see that f is not continuous. For, if V open in the uniform topology has preimage that is not open in R, V is still open in the box topology and still has the same non-open preimage. Next, by Example 19.2, we see that g is not continuous. Last, for h, we choose B = (−1, 1) × (− 212 , 212 ) × (− 312 , 312 ) × · · · , and suppose its preimage h−1 (B) is open. This implies h((−δ, δ)) ⊆ B, and so applying πn gives hn ((−δ, δ)) = (− nδ , nδ ) ⊆ (− n12 , n12 ) for all n, a contradiction. Solution for (b). We note that since the product topology is Hausdorff by Theorem 19.4 and both the uniform and box topologies are finer than the product topology by Theorems 19.1 and 20.4, if a sequence converges to a point p in one topology, it must converge to the same point in the finer topologies. For, if the sequence converges to q in the finer topology, then it also converges to q in the coarser topology, and by the Hausdorff property p = q. Q Consider wn . For the product topology, we recall that any basic open set U = Uα 3 0 is the product of finitely many open subsets of R with infinitely many copies of R. Letting N be the largest α such that Uα ( R, we see that wn ∈ U for all n > N since the first N components are zero, and the rest are trivially in the remaining copies of R of U . Thus, wn → 0 in the product topology. Now we only have to check if the sequence converges to zero in the other topologies by the above. In the uniform topology, ρ(wn , 0) = 1 for all n, and so the sequence does not converge. For, if we choose any ball U = B(0, r) ⊆ Rω for r < 1, wn ∈ / U for all n. Finally, since the box topology is finer than the uniform topology by Theorem 20.4, we see that this same open set U is such that wn ∈ / U for all n, and so wn does not converge in the box topology, either. Consider xn and yn . We claim they both converge to zero in the uniform topology. For any open set 0 ∈ U ⊆ Rω in the uniform topology, we can find B(0, ) such that B(0, ) ⊆ U ; then, we can find N such that 1/N < . We then see that xn , yn ∈ B(0, ) ⊆ U for all n ≥ N , and so xn , yn → 0 in the uniform topology. Moreover, since the uniform topology is finer than the product topology, we see that this implies 11

xn , yn → 0 in the product topology as well. For the box topology, though, Q∞ we see that neither sequence converges. For, we can construct the set 0 ∈ U = n=1 (−1/n, 1/n) (where we only consider sets containing zero by the above), which does not contain xn , yn for any n. Q For zn , we see that for any open set 0 ∈ U = Uα ⊆ Rω in the box topology, for N large enough 1/n ∈ U1 , U2 for n ≥ N , and so zn ∈ U for all n ≥ N , since by hypothesis 0 ∈ U , the third component onwards of zn are always in their respective Uα . Thus, zn → 0 in the box topology; since the box topology is finer than both the uniform and product topologies, we see that this implies zn converges in the other two topologies as well. Exercise 20.5. Let R∞ be the subset of Rω consisting of all sequences that are eventually zero. What is the closure of R∞ in Rω in the uniform topology? Justify your answer. Solution. We claim that A = R∞ is the set of all sequences that converge to zero; we denote this latter set by X. It suffices to show by Theorem 17.5 that x ∈ X if and only if every basis element U 3 x intersects A. First suppose x ∈ X and let U 3 x be a basis element in the uniform topology; we then see that we can find an open ball B(x, ) ⊆ U . We know we can find N ∈ N such that |xn | < for all n ≥ N by the definition of convergence. Then, define y such that yn = xn for all n < N , and zero otherwise; this means y ∈ A. Then, ρ(x, y) < , and so y ∈ B(x, ) ∩ A. Now suppose x ∈ / X; it suffices to find a basis element containing x that does not intersect A. Since x ∈ / X, there exists a ball B(0, ) ⊆ R such that {xn }n≥N 6⊆ B(0, ) for any N . The ball B(x, /2), then, does not intersect A, since for any y ∈ B(x, /2), it is not the case that yn = 0 for all n ≥ N for some N . Exercise 20.6. Let ρ be the uniform metric on Rω . Given x = (x1 , x2 , . . .) ∈ Rω and given 0 < < 1, let U (x, ) = (x1 − , x1 + ) × · · · × (xn − , xn + ) × · · · . (a) Show that U (x, ) is not equal to the -ball Bρ (x, ). (b) Show that U (x, ) is not even open in the uniform topology. (c) Show that [ Bρ (x, ) = U (x, δ). δ<

Proof of (a). Consider the point y = (yn )n∈N ,

n X 1 1 yn = xn + = xn + 1 − n . 2k 2 k=1 12

yn ∈ (xn − , xn + ) for all n implies y ∈ U (x, ), while y ∈ / Bρ (x, ) since ρ(x, y) = sup d(xn , yn ) = . n

Proof of (b). U (x, ) is not open since the point y in (a) has no neighborhood contained in Bρ (x, ). For, suppose Bρ (y, δ) ⊆ U (x, ). We can find N such that ∞ X δ 1 > , 2 2k k=N +1

P since 1/2k converges, and so its tail becomes infinitesimally small. We see that 0 6= N and yN = yN + δ/2, y0 ∈ Bρ (y, δ) then, defining y0 such that yn0 = yn for all nP ∞ 1 0 0 but y ∈ / U (x, ) since yn = yn +δ/2 > yn + k=N +1 2k = xn +, a contradiction. Proof of (c). The ⊇ direction is clear, since each U (x, δ) ⊆ Bρ (x, ) by the fact that δ < . Now suppose z ∈ BρS (x, ); if ρ(x, z) = ξ, then we can find δ ∈ (ξ, ) so that z ∈ U (x, ξ), i.e., Bρ (x, ) ⊆ δ< U (x, δ). Exercise 20.8. Let X be the subset of Rω consisting of all sequences x such that P 2 xi converges. Then the formula ∞ X d(x, y) = (xi − yi )2

"

#1/2

i=1

defines a metric on X. On X we have the three topologies it inherits from the box, uniform, and product topologies on Rω . We have also the topology given by the metric d, which we call the `2 -topology. (a) Show that on X, we have the inclusions box topology ⊃ `2 -topology ⊃ uniform topology. ∞ (b) The set R of all sequences that are eventually zero is contained in X. Show that the four topologies that R∞ inherits as a subspace of X are all distinct. (c) The set Y H= [0, 1/n] n∈Z+

is contained in X; it is called the Hilbert cube. Compare the four topologies that H inherits as a subspace of X.

13

Proof of (a). box topology ⊃ `2 -topology. Let U 3 x be a basis element in the `2 topology. Then, there exists U = Bd (x, ) ⊆ U of the `2 -topology. Q a basis element i/2 We claim that V = X ∩ (xi − /2 , xi + /2i/2 ) 3 x basis element of the box topology is contained in U . Suppose y ∈ V ; then ∞ ∞ X X 2 2 d(x, y) = = 2 =⇒ y ∈ U, (xi − yi ) < i 2 i=1 i=1 2

i.e., V ⊆ U ⊆ U, and box topology ⊃ `2 -topology by Lemma 13.3. `2 -topology ⊃ uniform topology. Let U 3 x be a basis element in the uniform topology. Then, there exists a basis element U = X ∩ Bρ (x, ) ⊆ U of the uniform topology. We claim V = Bd (x, ) basis element of the `2 -topology is contained in U . If > 1, then trivially V ⊆ U = X, so we assume ≤ 1. Suppose y ∈ V ; then ρ(x, y) = sup |xi − yi | ≤ d(x, y) < =⇒ y ∈ U, i.e., V ⊆ U ⊆ U, and `2 -topology ⊃ uniform topology by Lemma 13.3. Proof of (b). By (a) and Theorem 20.4, we have the inclusions box topology ⊃ `2 -topology ⊃ uniform topology ⊃ product topology, since if T1 ⊃ T2 in the ambient space, we would also have T10 ⊃ T20 , where Ti0 are the subspace topologies induced by Ti on X, by the fact that the open sets in the subspace X are the open sets of the ambient space intersected with X. We claim these are strict inclusions. Q box topology ) `2 -topology. Consider the open set U = R∞ ∩ (−1/i, 1/i) 3 0 in the box topology. Consider any neighborhood V 3 0; then, there exists some V = R∞ ∩ Bd (0, ) for > 0 contained in V. We can find N ∈ N such that 1/N < , and so x such that xi = 0 for all i except xN = 1/N is contained in V since d(0, x) = 1/N < and therefore V, but not in U . Hence, U is open in the box topology but not in the `2 -topology by p. 78, and so box topology ) `2 -topology. `2 -topology ) uniform topology. Consider the open set U = R∞ ∩ Bd (0, 1) in the `2 -topology. Consider any neighborhood V 3 0; then, there exists some V = R∞ ∩ Bρ (0, ) for > 0 contained in V. We can find N ∈ N such that N 2 > 4, and so x such that xi = /2 for all 1 ≤ i ≤ N and 0 otherwise is contained in V and therefore V, yet 2

d(x, 0) =

∞ X i=1

x2i = N

2 > 1 =⇒ d(x, 0) > 1, 4 14

and so x ∈ / U . Hence, U is open in the `2 -topology but not in the uniform topology by p. 78, and so `2 -topology ) uniform topology. uniform topology ) product topology. Consider the open set U = R∞ ∩ Bρ (0, 1) Q in the uniform topology. Consider any V = R∞ ∩ Uα 3 0 where Uα = R for all but finitely many α. Let N be such that UN = R; then, x such that xi = 0 for all i except |xN | ≥ 1 is in V but not in U . Hence, U is open in the uniform topology but not in the product topology by p. 78, and so uniform topology ) product topology. Solution for (c). We claim that box topology ) (`2 -topology = uniform topology = product topology), i.e., the box topology is strictly finer than the other topologies, which are equal. To show the equality, it suffices to show product topology ⊃ `2 -topology, for then we have `2 -topology ⊃ uniform topology ⊃ product topology ⊃ `2 -topology by the same argument as in (b), and so we must have equality throughout. So, consider U 3 x open in the `2 -topology; then, we can find a basis element U = H∩Bd (x, ) ⊆ U of the `2 -topology for some > 0. Let δ = /[ζ(2) + 1]1/2 , and choose N such that P ∞ 2 2 i=N 1/i < δ , which exists since |ζ(2)| < ∞. We claim that "N −1 # Y ∞ Y δ δ V =H∩ xi − , xi + × R 3 x, i i i=1 i=N basis element of the product topology is contained in U . Suppose y ∈ V ; then ∞ N X 1 X1 + < δ 2 [ζ(2) + 1] = 2 =⇒ y ∈ U, d(x, y) = (xi − yi ) < δ 2 2 i i i=1 i=1 i=N 2

∞ X

2

2

i.e., V ⊆ U ⊆ U, and product topology ⊃ `2 -topology by Lemma 13.3. Thus, we have the equality desired. It remains to show the box topology is strictly finer than the other topologies; 2 since the other topologies are Q equal it suffices to show box topology ) ` -topology. But the open set U = H ∩ (−1/i, 1/i) 3 0 is open in the box topology but not the `2 -topology by the same argument as the proof that box topology ) `2 -topology in (b), and so we are done.

21

The Metric Topology (continued)

Exercise 21.1. Let A ⊆ X. If d is a metric for the topology of X, show that d|A×A is a metric for the subspace topology on A. 15

Proof. Clearly d|A×A is a metric since it inherits all the properties for a metric from the metric d for X; it therefore suffices to show that every basis element for the subspace topology on A contains some open ball defined by d|A×A , and vice versa, by Lemma 13.3. So, suppose B is a basis element for the metric topology on A; B = Bd|A×A (x, r) for some x ∈ A and r ∈ R. But then, B = Bd (x, r) ∩ A by definition, and so the subspace topology is finer than the metric topology. Conversely, suppose B is a basis element for the subspace topology on A; it equals Bd (x, r) ∩ A for some basis element Bd (x, r) of X. Let y ∈ Bd (x, r) ∩ A; we see that the set Bd|A×A (y, r − d(x, y)) ⊆ A ∩ Bd (x, r) is a basis element for the metric topology contained in A ∩ Bd (x, r), since z ∈ Bd|A×A (y, r − d(x, y)) is such that d(z, x) ≤ d(z, y) + d(x, y) = d|A×A (z, y) + d(x, y) < r − d(x, y) + d(x, y) = r. Thus, the metric topology is finer than the subspace topology. Combining the two inclusions, we see the topologies are equal. Exercise 21.2. Let X and Y be metric spaces with metrics dX and dY , respectively. Let f : X → Y have the property that for every pair of points x1 , x2 of X, dY (f (x1 ), f (x2 )) = dX (x1 , x2 ). Show that f is an imbedding. It is called an isometric imbedding of X in Y . Proof. We first show f is injective: f (x1 ) = f (x2 ) =⇒ dY (f (x1 ), f (x2 )) = 0 =⇒ dX (x1 , x2 ) = 0 =⇒ x1 = x2 by properties of metrics, and so we have an injective map. Now we consider the map f 0 : X → im(X) ⊆ Y , which is a bijection; it suffices to show that f 0 , f 0−1 are continuous to show f is an imbedding. Let x ∈ X and > 0 be given; then, letting δ = , we have dX (x, y) < δ =⇒ dY (f 0 (x), f 0 (y)) < , and so f is continuous. Given y ∈ Y and > 0, letting δ = gives dY (x, y) = dY (f (a), f (b)) < δ =⇒ dX (f 0−1 (x), f 0−1 (y)) = dX (a, b) < , where a, b exist by the bijectivity of f , and so f 0−1 is continuous. Exercise 21.3. Let Xn be a metric space with metric dn , for n ∈ Z+ . 16

(a) Show that ρ(x, y) = max{d1 (x1 , y1 ), . . . , dn (xn , yn )} is a metric for the product space X1 × · · · × Xn . (b) Let di = min{di , 1}. Show that D(x, y) = sup{di (xi , yi )/i} Q is a metric for the product space Xi . Proof of (a). ρ satisfies properties (1) and (2) on p. 119 since the components do; it then suffices to show the triangle inequality. We first have di (xi , zi ) ≤ d(xi , yi ) + d(yi , zi ) for all i. Then, by definition of ρ, di (xi , zi ) ≤ ρ(x, y) + ρ(y, z) for all i. But since this is true for all i, we have that ρ(x, z) ≤ ρ(x, y) + ρ(y, z). Q We now show that this defines a metric for the product space. First let B = Ui Q be a basis element of Xi , and let x ∈ B. For each i, there is an i such that Bdi (xi , i ) ⊆ Ui . Choosing = min{1 , . . . , n }, we see that Q Bρ (x, ) ⊆ B, since if y ∈ Bρ (x, ), di (xi , yi ) ≤ ρ(x, y) < ≤ i , and so y ∈ Ui as desired. Thus the metric topology is finer than the product topology. Conversely, let Bρ (x, ) be a basis element in the metric topology; since it is the product Bdi (xi , ), we see that the product topology is finer than the metric topology. These two facts imply the two topologies are equal. Proof of (b). D satisfies properties (1) and (2) on p. 119 since the components do; it then suffices to show the triangle inequality. We first have di (xi , zi ) di (xi , yi ) di (yi , zi ) ≤ + ≤ D(x, y) + D(y, z) i i i for all i. But since this is true for all i, we have that di (xi , zi ) D(x, z) = sup ≤ D(x, y) + D(y, z). i We now show that this defines a metric for the product space. Let U be open in the metric topology and let x ∈ U ; choose an open ball BD (x, ) ⊆ U . Choose N such that 1/N < , and let V = Bd1 (x1 , ) × · · · × BdN (xN , ) × XN +1 × XN +2 × · · · . Q We claim V ⊆ BD (x, ) ⊆ U . Given y ∈ Xi , di (xi , yi )/i ≤ 1/N for i ≥ N . Therefore, d1 (x1 , z1 ) dN (xN , zN ) 1 D(x, y) ≤ max ,··· , , . 1 N N 17

If y ∈ V , this expression is less than , so V ⊆ BD (x, ) as desired, and the product topology is finer than theQmetric topology. Conversely, let U = Ui where Ui is open in Xi for α1 , . . . , αn and Ui = Xi otherwise. Let x ∈ U be given, and choose Bdi (xi , i ) ⊆ Xi for i = α1 , . . . , αn , where each i ≤ 1. Then, defining = min{i /i | i = α1 , . . . , αn }, we claim that x ∈ BD (x, ) ⊆ U . Let y be a point of BD (x, ). Then, for all i, di (xi , yi ) ≤ D(x, y) < . i Now if i = α1 , . . . , αn , then Q ≤ i /i, so that di (xi , yi ) < i ≤ 1; it follows that |xi − yi | < i , and so y ∈ Ui as desired. We thus have that the metric topology is finer than the product topology; combined with the above this implies the topologies are equal.

22

The Quotient Topology

Exercise 22.2. (a) Let p : X → Y be a continuous map. If there is a continuous map f : Y → X such that p ◦ f equals the identity map of Y , then p is a quotient map. (b) If A ⊂ X, a retraction of X onto A is a continuous map r : X → A such that r(a) = a for each a ∈ A. Show that a retraction is a quotient map. Proof of (a). If V ⊆ Y with U = p−1 (V ) ⊆ X open, f −1 (U ) = f −1 (p−1 (V )) = (p ◦ f )−1 (V ) = V is open. Thus, p is a quotient map. Proof of (b). Let ι : A → X be the inclusion map; then, r ◦ ι is the identity on A, hence r is a quotient map by (a). Exercise 22.4. (a) Define an equivalence relation on the plane X = R2 as follows: x0 × y0 ∼ x1 × y1

if x0 + y02 = x1 + y12 .

Let X ∗ be the corresponding quotient space. It is homeomorphic to a familiar space; what is it? (b) Repeat (a) for the equivalence relation x0 × y0 ∼ x1 × y1

18

if x20 + y02 = x21 + y12 .

Solution for (a). Set g(x × y) = x + y 2 ∈ R. We see it is a surjection onto R since R × {0} 7→ R. It is continuous since for x0 × y0 ∈ X, given > 0, letting δ = min(1, /2(|y0 | + 1)), ρ(x0 × y0 , x × y) < δ implies |g(x0 × y0 ) − g(x × y)| = |(x0 + y02 ) − (x + y 2 )| ≤ |x0 − x| + |y02 − y 2 | ≤ |x0 − x| + |y0 + y||y0 − y| ≤ |x0 − x| + |y − y0 |2 + 2|y0 ||y0 − y| < 2(|y0 | + 1)δ < . If we define f : R → X by x 7→ x × 0, which is continuous since (a, b) × (c, d) maps back to (a, b) which is open in R, we see g◦f is the identity on R, and so g is a quotient map by the lemma above. Since x0 × y0 ∼ x1 × y1 ⇐⇒ g(x0 × y0 ) = g(x1 × y1 ), by Corollary 22.3, this induces a bijective continuous map g 0 : X ∗ → R, which is a homeomorphism since g was a quotient map. Solution for (b). Set g(x × y) = x2 + y 2 ∈ R. We see it is a surjection onto R≥0 , since R × {0} 7→ R≥0 , and it does not map to anything else since x2 + y 2 ≥ 0 for all x, y. It is continuous since for x0 ×y0 ∈ X, given > 0, letting δ = min(1, /2(|y0 |+|x0 |+1)), ρ(x0 × y0 , x × y) < δ implies |g(x0 × y0 ) − g(x × y)| = |(x20 + y02 ) − (x2 + y 2 )| ≤ |x20 − x2 | + |y02 − y 2 | ≤ |x0 − x||x − x0 + 2x0 | + |y0 − y||y − y0 + 2y0 | ≤ |x0 − x|(1 + 2|x0 |) + |y0 − y|(1 + 2|y0 |) ≤ 2δ(|x0 | + |y0 | + 1) < . √ We define f : R≥0 → X by x 7→ x × 0, which is continuous since the preimage of (a, b) × (c, d), if (c, d) 3 0, is the open set R≥0 ∩ (a0 , b0 ), where a0 = a2 if a ≥ 0, and −1 otherwise, and similarly for b0 (we chose −1 out of convenience; we really only have to make sure the preimage is a half-open set [0, b0 ) or the empty set in these cases); if (c, d) 63 0, then the preimage would be empty. We then see g ◦ f is the identity on R≥0 , and so g is a quotient map by the lemma above. Since x0 × y0 ∼ x1 × y1 ⇐⇒ g(x0 × y0 ) = g(x1 × y1 ), by Corollary 22.3, this induces a bijective continuous map g 0 : X ∗ → R≥0 , which is a homeomorphism since g was a quotient map. Exercise 22.6. Recall that RK denotes the real line in the K-topology. Let Y be the quotient space obtained by RK by collapsing the set K to a point; let p : RK → Y be the quotient map. 19

(a) Show that Y satisfies the T1 axiom, but is not Hausdorff. (b) Show that p × p : RK × RK → Y × Y is not a quotient map. Proof of (a). Recall by p. 141 that it suffices to show every element in the partition, i.e., one-point sets {x} for x ∈ / K and K itself, are closed in RK . The former are closed since RK is T1 since it is Hausdorff by Example 31.1, and the latter is closed since it is the complement of R \ K. Thus, Y is T1 . We now show Y is not Hausdorff. We claim that p(0), p(K) are not separable; note p(0) 6= p(K) since they are in different equivalence classes. Suppose Y is Hausdorff, and let V1 3 p(0), V2 3 p(K) be a separation in Y ; they have open preimages U1 = p−1 (V1 ) 3 0, U2 = p−1 (V2 ) ⊇ K by definition of a quotient map. There then exists (a, b) \ K 3 0 contained in U1 , and choosing n ∈ N such that 1/n < b, there exists (c, d) 3 1/n contained in U2 , where we can assume 1/(n + 1) ≤ c, since if not, we can take the intersection with (1/(n + 1), d). Then, (c, 1/n) ⊆ U1 ∩ U2 , and so p((c, 1/n)) ⊆ V1 ∩ V2 , which is a contradiction, and so Y is not Hausdorff. Proof of (b). By Exercise 17.13, we see that since Y is not Hausdorff by (a), the diagonal ∆Y ⊆ Y × Y is not closed. (p−1 × p−1 )(∆Y ) = ∆K ∪ (K × K), where ∆K ⊆ RK × RK is the diagonal in RK . However, ∆K is closed by Exercise 17.13 since RK is Hausdorff by Example 31.1, and so ∆K ∪ (K × K) is closed since is is the finite union of closed sets. Thus, the inverse image of the non-closed set ∆Y is closed, and so p × p is not a quotient map.

3

Connectedness and Compactness

23

Connected Spaces

Exercise 23.8. Determine whether or not Rω is connected in the uniform topology. Solution. Let Rω = A ∪ B, where A is the set of bounded sequences and B is the set of unbounded sequences of reals. A, B are disjoint, and so it remains to show they are open. Suppose a = (a1 , a2 , . . .) ∈ A and b = (b1 , b2 , . . .) ∈ B. Since |ai | < N for all i for some N , and since |bi | > N + 1 for all i larger than some I, we have that d(ai , bi ) = 1 for all i ≥ I. Thus, ρ(a, b) = 1 for any a ∈ A, b ∈ B, and so the open balls with radius 1/2 around a, b are fully contained in A, B respectively. Exercise 23.11. Let p : X → Y be a quotient map. Show that if each set p−1 ({y}) is connected, and if Y is connected, then X is connected.

20

Proof. Suppose not. Then, X = A ∪ B for A, B open, disjoint sets. Consider C = {y ∈ Y | p−1 ({y}) ⊆ A}, D = {y ∈ Y | p−1 ({y}) ⊆ B}; we see that these sets are such that C ∪ D = Y since p−1 ({y}) connected implies it is in either A or B by Lemma 23.2. C, D are then disjoint by definition and p−1 (C) = A, p−1 (D) = B by the fact that p is surjective. p quotient map implies that C, D are then open, and so Y = C ∪ D is a separation, a contradiction.

24

Connected Subspaces of the Real Line

Exercise 24.7. (a) Let X and Y be ordered sets in the order topology. Show that if f : X → Y is order preserving and surjective, then f is a homeomorphism. (b) Let X = Y = R+ . Given a positive integer n, show that the function f (x) = xn is order preserving and surjective. Conclude that its inverse, the nth root function, is continuous. (c) Let X be the subspace (−∞, −1)∪[0, ∞) of R. Show that the function f : X → R defined by setting f (x) = x + 1 if x < −1, and f (x) = x if x ≥ 0, is order preserving and surjective. Is f a homeomorphism? Compare with (a). Proof of (a). f is injective since if f (a) = f (b) but a 6= b, then (with possible swapping) a < b, and so f (a) < f (b), a contradiction. We thus must show f and f −1 are continuous. But f is continuous since f −1 ((a, b)) = (f −1 (a), f −1 (b)) is open (apply the same argument to the intervals of the form [a0 , b), (a, b0 ] for a0 , b0 minimal and maximal respectively); the same argument applies for f −1 as well. Proof of (b). f (x) = xn is order preserving since a < b =⇒ a/b < 1 =⇒ an /bn < 1 =⇒ an < bn =⇒ f (a) < f (b). f is continuous since it is the product of n copies of the identity function, which is continuous. We want to show f is surjective. Letting N = {xn | x ∈ Z≥0 }, we see that every real number y ∈ Y is between two consecutive members of N , or it is already an nth power of an integer, in which case it is trivially mapped to by its nth root. In the case y ∈ Y is not an nth power, we have f (n) < y < f (n + 1), and so by the Intermediate value theorem (Theorem 24.3), we see that there exists a point c ∈ X such that f (c) = r, i.e., f is surjective. Since f is order preserving and surjective, by (a) it is then a homeomorphism, and so f −1 , the nth root function, is also continuous. Proof of (c). f is order-preserving on (−∞, −1) since a < b =⇒ f (a) = a + 1 < b + 1 = f (b), and on [0, ∞) since it is the identity. We check that it is order preserving around the boundary. So, suppose a < −1 and b ≥ 0. Then, a < b but 21

also a+1 < 0 ≤ b, and so f is order-preserving. f is surjective since if x ∈ R, if x ≥ 0 its preimage is itself, and if x < 0, its preimage is x − 1. f is not a homeomorphism by Theorem 23.6 since R is connected but X is not, by considering f −1 (R). This does not contradict (a) since X is not in the order topology. Even if R is in the order topology, the subspace topology induced on X is not the order topology. For, (−1/2, 1) is open in R, and so (−1/2, 1) ∩ X = [0, 1) is open in X, but not open in the order topology on X. Exercise 24.8. (a) Is a product of path-connected spaces necessarily path connected? (b) If A ⊂ X and A is path connected, is A necessarily path connected? (c) If f : X → Y is continuous and X is path connected, is f (X) necessarily path connected? T (d) If S {Aα } is a collection of path-connected subspaces of X and if Aα 6= ∅, is Aα necessarily path connected? Q Solution for (a). Yes. Let X = Xα , x, y ∈ X. Since each Xα is path connected, we have fα : [0, 1] → Xα continuous such that fα (0) = xα , fα (1) = yα , where we assume the closed interval is [0, 1] after composition with multiplication and addition, which are continuous operations. Thus we have the function f = (fα ), which is continuous by Theorem 19.6, with f (0) = x, f (1) = y, and so X is path-connected. Solution for (b). No, since S in Example 24.7 is not path-connected while S is: it is the image of the continuous map x 7→ (x, sin(1/x)) from R>0 to R2 . Solution for (c). Yes. For, let x, y ∈ f (X), and choose x0 ∈ f −1 (x), y0 ∈ f −1 (y). Then, there exists continuous g : [a, b] → X such that g(a) = x0 , g(b) = y0 , and so its composition f ◦ g : [a, b] → Y is continuous with (f ◦ g)(a) = x, (f ◦ g)(b) = y. S T Solution for (d). Yes. SLet x, y ∈ Aα and p ∈ Aα . Then, there exists a continS uous map f : [a, b] → Aα with f (a) = x, f (b) = p, and similarly g : [b, c] → Aα with f (b) = p, f (c) = y, since a, p ∈ Aα for some α and similarly for y (we are free to have Dom g = [b, c] by composition with multiplication and addition, which are continuous). Then, by the pasting lemma (Theorem 18.3) since f, g are continuous and f (b) = g(b), we see that h = f on [a, b] and h = g on [b, c] is a continuous map such that h(a) = x, h(c) = y. Exercise 24.12. Recall that SΩ denotes the minimal uncountable well-ordered set. Let L denote the ordered set SΩ × [0, 1) in the dictionary order, with its smallest element deleted. The set L is a classical example in topology called the long line. 22

Theorem. The long line is path connected and locally homeomorphic to R, but it cannot be imbedded in R. (a) Let X be an ordered set; let a < b < c be points of X. Show that [a, c) has the order type of [0, 1) if and only if both [a, b) and [b, c) have the order type of [0, 1). (b) Let X be an ordered set. Let x0 < x1 < · · · be an increasing sequence of points of X; suppose b = sup{xi }. Show that [x0 , b) has the order type of [0, 1) if and only if each interval [xi , xi+1 ) has the order type of [0, 1). (c) Let a0 denote the smallest element of SΩ . For each element a of SΩ different from a0 , show that the interval [a0 × 0, a × 1) of SΩ × [0, 1) has the order type of [0, 1). (d) Show that L is path connected. (e) Show that every point of L has a neighborhood homeomorphic with an open interval in R. (f ) Show that L cannot be imbedded in R, or indeed in Rn for any n. Proof of (a). We first note order-preserving maps are injective. Letting f : A → B be such a map, if a1 , a2 ∈ A, then one is larger than the other by the comparability property of order relations, so one of f (a1 ), f (a2 ) is larger than the other, hence unequal. Now suppose [a, c) has order type [0, 1), and let f : [0, 1) → [a, c) be the order isomorphism. We claim g(x) = f {[f −1 (b)]x},

h(x) = f {f −1 (b) + [1 − f −1 (b)]x}

define order isomorphisms g : [0, 1) → [a, b) and h : [0, 1) → [b, c). They are orderpreserving since if x, y ∈ [0, 1), x < y =⇒ [f −1 (b)]x < [f −1 (b)]y =⇒ g(x) = f {[f −1 (b)]x} < f {[f −1 (b)]y} = g(y) x < y =⇒ f −1 (b) + [1 − f −1 (b)]x < f −1 (b) + [1 − f −1 (b)]y =⇒ h(x) = f {f −1 (b) + [1 − f −1 (b)]x} < f {f −1 (b) + [1 − f −1 (b)]y} = h(y) where the first implications are due to our linear transformations being strictly monotonic increasing, and the second since f is order-preserving. This also implies injectivity by the above. It remains to show surjectivity. Let z ∈ [a, b), z 0 ∈ [b, c). Then, 0 ≤ f −1 (z) < f −1 (b) and f −1 (b) ≤ f −1 (z 0 ) < 1, and so −1 −1 0 f (z ) − f −1 (b) f (z) g −1 = z, h = z0. f (b) 1 − f −1 (b) 23

Conversely, suppose [a, b) and [b, c) have order type [0, 1), and let g : [0, 1) → [a, b), h : [0, 1) → [b, c) be the order isomorphisms. We claim ( g(2x) if 0 ≤ x < 1/2 f (x) = h(2x − 1) if 1/2 ≤ x < 1 is an order isomorphism. It preserves orders since g, h preserve orders on their respective domains, and since if x < 1/2 ≤ y, applying f gives f (x) = g(2x) < b ≤ h(2y − 1) = f (y). This also shows injectivity by the above. f is surjective since if z ∈ [a, c), z < b =⇒ f [g −1 (z)/2] = z,

z ≥ b =⇒ f {[h−1 (z) + 1]/2} = z.

Proof of (b). Suppose [x0 , b) has order type [0, 1). For any i ∈ Z+ , by (a), [xi , b) has order type [0, 1); applying (a) again gives that [xi , xi+1 ) has order type [0, 1). Now suppose every [xi , xi+1 ) has order type [0, 1). If fi : [0, 1) → [xi , xi+1 ) are order isomorphisms, first define f : [0, ∞) → [x0 , b),

x 7→ fi (x − i) if x ∈ [i, i + 1),

which is well-defined since any x ∈ [0, ∞) is in some set of the form [i, i + 1). We claim f is an order isomorphism. If x, y ∈ [0, ∞), then x ∈ [i, i + 1), y ∈ [j, j + 1) for some i, j. Suppose x < y. Then, i 6= j =⇒ f (x) = fi (x − i) < xi+1 ≤ xj ≤ fj (y − j) = f (y), i = j =⇒ f (x) = fi (x − i) < fi (y − i) = f (y), since the fi are order-preserving; this also S implies injectivity by the above. To show surjectivity, we first know f maps onto i [xi , xi+1 ) by definition. So let z ∈ [x0 , b). Since b is the least upper bound of the {xi }, z is not an upper bound, so there exists i such that z ∈ [xi , xi+1 ). But then, since the fi are bijective as well, f (fi−1 (z)+i) = z. Now let g : [0, 1) → [0, ∞) be defined as x 7→ x/(1 − x); this is an order isomorphism since it has inverse x/(1 + x), and since it is strictly monotonic increasing. Thus, f ◦ g : [0, 1) → [x0 , b) is a bijection, and preserves orientation since f, g do. Proof of (c). Let a > a0 . We proceed by transfinite induction. SΩ is a well-ordered set, and so if we let J be the set of a ∈ SΩ such that the claim holds, it suffices to show that for every a ∈ J, Sa ⊆ J =⇒ a ∈ J. 24

We first show that either a has an immediate predecessor or there exists a sequence {ai } ⊆ Sa such that a = sup{ai }. Suppose a does not have an immediate predecessor. Then, we have the section Sa = {bi }, which is countable by definition of SΩ . (b1 , a] 6= ∅ since a has no immediate predecessor, and so let a1 ∈ (b1 , a]. We construct the ai inductively as follows: if we have an , let an+1 ∈ (sup{an , bn+1 }, a], which is nonempty as above. We then get a sequence of elements a1 < a2 < · · · < a. But since an > bn for all n by construction, we see that a ≥ sup{ai }. Moreover, if a > sup{ai }, then sup{ai } = bk for some k, for Sa contains all elements less than a, and hence sup{ai } < ak , contradicting that sup{ai } is an upper bound. Now suppose Sa ⊆ J. If a has an immediate predecessor a−1, then [a0 ×0, a×1) = [a0 × 0, (a − 1) × 1) ∪ [a × 0, a × 1) has order type [0, 1) by (a), for we have the order isomorphism [a × 0, a × 1) → [0, 1) defined by a × x 7→ x, which is trivially bijective and order-preserving since SΩ × [0, 1) was constructed with the dictionary order. On the other hand, if a does not have an immediate predecessor, then there exists a sequence {ai } ⊆ Sa such that a = sup{ai }, and so the claim follows by (b). Proof of (d). Let a × b, a0 × b0 be two points in L; suppose without loss of generality that a × b < a0 × b0 . By (c), [a0 × 0, a × 1) and [a0 × 0, a0 × 1) have order type [0, 1); by 9a), this implies [a0 ×0, a×b) and [a0 ×0, a0 ×b0 ) have order type [0, 1). Hence, by (a), Y = [a × b, a0 × b0 ) has order type [0, 1). Let f : [0, 1) → Y be the order isomorphism. We claim f is continuous. First, since Y is an interval, it is convex, and so by Theorem 16.4 the order topology on Y is the same as the subspace topology on Y inherited from L. Then, for any basis set A = (c × d, c0 × d0 ) ⊆ Y , f −1 (A) = (f −1 (c × d), f −1 (c0 × d0 )) since f is an order isomorphism, and moreover this preimage is open. Also, for any basis set B = [a × b, c0 × d0 ) ⊆ Y , f −1 (B) = [f −1 (a × b), f −1 (c0 × d0 )), which is again open. Thus, f is continuous. Finally, if we define ( f (x) if x ∈ [0, 1) F (x) = 0 0 a × b if x = 1 we have a continuous path F : [0, 1] → [a × b, a0 × b0 ] by the pasting lemma (Theorem 18.3), and so L is path connected. Proof of (e). Let a × b be a point in L. Since SΩ × [0, 1) does not have a maximal element, there is some a0 × b0 > a × b. Now by (c), there exists an order isomorphism f : [0, 1) → [a0 × 0, a0 × b0 ). Restricting f to (0, 1), we get another order isomorphism f : (0, 1) → (a0 × 0, a0 × b0 ). The set [a0 × 0, a0 × b0 ) is open in SΩ × [0, 1), and so (a0 × 0, a0 × b0 ) is open in L, and is a neighborhood of a × b. We claim (a0 × 0, a0 × b0 ) is homeomorphic to (0, 1). We already have a bijection that is continuous by the same argument as in (d), and so it suffices to show f is 25

open as well. But if (x, y) ⊆ (0, 1) is a basis set, then f (x, y) = (f (x), f (y)) since f is an order isomorphism, and moreover open since the topology on (a0 × 0, a0 × b0 ) is the order topology. Proof of (f ). Suppose L could be imbedded in Rn ; then, every subspace of Rn has a countable basis by Example 30.1, and since L is homeomorphic with such a subspace, it also has a countable basis. Now, since X = (SΩ × {0}) \ {a0 × 0} is a convex subset of L, the subspace topology on X is the same as the order topology by Theorem 16.4. Thus, the intersection of the countable basis for L with X forms a countable basis by Theorem 30.2. This implies that there is a countable subset Y of X that is dense in X by Theorem 30.3. By Theorem 10.3, though, this subset Y has an upper bound x in X. Thus, ∅ 6= (x, Ω) ⊆ X \ Y , and so the closure of Y is not all of X, a contradiction.

25

Components and Local Connectedness

Exercise 25.2. (a) What are the components and path components of Rω (in the product topology)? (b) Consider Rω in the uniform topology. Show that x and y lie in the same component of Rω if and only if the sequence x − y = (x1 − y1 , x2 − y2 , . . .) is bounded. (c) Give Rω the box topology. Show that x and y lie in the same component of Rω if and only if the sequence x − y is “eventually zero.” Solution for (a). By Exercise 24.8(a), Rω is path connected, for Theorem 19.6 is not limited to finite product topologies. Thus, Rω is the only path component, and so Rω is the only component as well since path connected =⇒ connected. Proof of (b). We first define ϕ : x 7→ x − y. We recall that since ρ(ϕ(x), ϕ(z)) = ρ(x, z), by Exercise 21.2 ϕ is an isometric imbedding that is moreover surjective (the preimage of any z is z + y), ϕ is a homeomorphism. Thus, x − y is in the same component as 0 if and only if x is in the same component as y, for ϕ, ϕ−1 do not modify the topology of Rω . It therefore suffices to check the case y = 0. Suppose x is bounded; then, we define f : [0, 1] → Rω where f (t) = (x1 t, x2 t, . . .). This is continuous since given > 0, B(f (t), ) ⊇ f (B(t, / sup{|xn |})), where sup{|xn |} < ∞ by boundedness

26

of x. Thus, f connects 0 and x, i.e., they are in the same path component, and therefore the same component by Theorem 25.5. Conversely, recall by Exercise 23.8 that we have the separation Rω = A ∪ B, where A is the set of bounded sequences and B is the set of unbounded sequences of reals. If x is unbounded it is in B and so is not in the same component as 0. Proof of (c). x “eventually zero” here means that xi = 0 for all i ≥ N for some N . Note by the same argument as in (b), it suffices to consider the case y = 0. Suppose first that x is not eventually zero. Define the function f = (fn ), where fn (a) = na/|xn | if xn 6= 0, and a otherwise. f is continuousQsince eachQfn is continuous Vn . Note that since it is linear, and so if fn−1 (Un ) = Vn , we have f −1 ( Un ) = −1 this is a bijection since each component has an inverse fn (a) = |xn |a/n if xn 6= 0, and a otherwise, and moreover since the inverse is continuous since each component is linear, we have a homeomorphism f : Rω → Rω . Since there are infinitely many n such that xn 6= 0, and so infinitely many n such that fn (xn ) = n, we have that f (x) is unbounded, and thus, by the separation of Rω in the box topology in Example 23.6, we have that f (x) and 0 are in different components. Since f is a homeomorphism, this implies x and 0 are in different components as well. Conversely, suppose x is eventually zero. Then, xn = 0 for all n ≥ N for some N , and so x ∈ RN × {0} × {0} × · · · ⊆ Rω ; this subspace is homeomorphic to RN . Since RN is connected by Theorem 23.6, we see that x and 0 are in the same component.

27

Compact Subspaces of the Real Line

Exercise 27.4. Show that a connected metric space having more than one point is uncountable. Proof. Let X be a connected metric space with the metric d, and let x0 , x1 ∈ X be distinct. Let d(x0 , x1 ) = r, and define f (x) = d(x0 , x). f is continuous by the discussion on p. 175. We see that f (x0 ) = 0, f (x) = r, and so by the intermediate value theorem (Theorem 24.3), f (X) ⊇ [0, r], i.e., f maps onto [0, r]. Now suppose X is countable. Then, by Theorem 7.1 there exists a surjective function g : Z+ → X, and so f ◦ g : Z+ → f (X) maps onto [0, r], which is a contradiction since [0, r] is uncountable by Corollary 27.8.

29

Local Compactness

Exercise 29.4. Show that [0, 1]ω is not locally compact in the uniform topology. 27

Proof. Suppose X = [0, 1]ω is locally compact, and in particular at 0. Then, there exists C compact that contains a neighborhood U 3 0. There exists X ∩Bρ (0, ) ⊆ U ; we see that {0, /3}ω ⊆ X ∩ Bρ (0, ). {0, /3}ω is closed since {0, /3}ω =

Y Y Y {0, /3} = {0, /3} = {0, /3} = {0, /3}ω

in the product topology by Theorem 19.5, which is finer than the uniform topology, and so it is compact by Theorem 26.2 since it is a closed subset of C compact, i.e., limit point compact by Theorem 28.2. We claim this is a contradiction. Consider x ∈ X, and the ball X ∩ Bρ (x, /9). Note that the distance between any two distinct points of {0, /3}ω is /3, and so since the diameter of X ∩ Bρ (x, /9) is at most 2/9, X ∩ Bρ (x, /9) contains at most one point of {0, /3}ω . Thus, {0, /3}ω contains no limit points, and so is not limit point compact, a contradiction. Exercise 29.8. Show that the one-point compactification of Z+ is homeomorphic to the subspace {0} ∪ {1/n | n ∈ Z+ } of R. Proof. Let K = {1/n | n ∈ Z+ }. Let f : R+ → R+ such that f (x) = 1/x; this is a homeomorphism since it is continuous and is its own inverse. By Theorem 18.2(d) and 18.2(e), f : Z+ → f (Z+ ) = K is continuous, and again is a homeomorphism since it is its own inverse. Now consider Y = {0} ∪ K, which is closed and bounded and therefore compact by Theorem 27.3, and Hausdorff by Theorem 17.11. Since K 0 = {0} by Example 17.8, we know Y is the one-point compactification of K. If X = {p} ∪ Z+ is the one-point compactification of Z+ , and letting g : p 7→ 0 ∈ Y , which is clearly continuous, the function h : X → Y defined by the pasting lemma (Theorem 18.3) applied to f, g is also continuous, and has continuous inverse defined by the pasting lemma applied to f −1 , g −1 , and so is a homeomorphism X ↔ Y .

4

Countability and Separation Axioms

30

The Countability Axioms

Exercise 30.4. Every compact metrizable space X has a countable basis. Proof. For given n ∈ Z+ , we have an open cover ofSX by B(x, 1/n) for each x ∈ X; since X is compact, let An be the finite subcover. n An is countable since it is the countable union of finite sets; we claim it is a basis for X. Let U ⊆ X open and x ∈ U . Since X is metrizable, there exists B(x, δ) ⊆ U for some δ > 0. Let N such 28

that 2/N < δ. Since AN covers X, there exists B(y, 1/N ) 3 x. B(y, 1/N ) ⊆ B(x, δ), for if we choose z ∈ B(y, 1/N S), d(x, z) ≤ d(x, y) + d(y, z) ≤ 1/N + 1/N < δ. Thus, x ∈ B(y, 1/N ) ⊆ U , and so n An is a countable basis by Lemma 13.2. Exercise 30.5. (a) Show that every metrizable space with a countable dense subset has a countable basis. (b) Show that every metrizable Lindelöf space has a countable basis. Proof of (a). Let X be a metrizable space and A a countable dense subset. We claim that the set of open balls in X below is a basis for X: B := {B(a, 1/n) ⊆ X | a ∈ A, n ∈ N}. Note B is countable since is in bijection with A × N. So let x be contained in an open subset U ⊆ X; since X is metrizable, x ∈ B(x, ) ⊆ U for some small . Let n be such that 1/n < /2. Then, since A is dense, some a ∈ A is contained in B(x, 1/n), and conversely x ∈ B(a, 1/n). By the triangle inequality, x ∈ B(a, 1/n) ⊆ U , so by Lemma 13.2 we are done. Proof of (b). Let X be a metrizable space. Then, the set of open balls Ben := {B(x, 1/n) ⊆ X | x ∈ X} is an open cover of X for each S n ∈ N; since X is Lindelöf, it has a countable subcover Bn . We claim B := n∈N Bn is a basis for X; note it is countable since it is a countable union of countable sets. So let x ∈ B(x, ) ⊆ U as before, and let n such that 1/n < /2. Then, there is some x0 ∈ X such that B(x0 , 1/n) ∈ Bn contains x. By the triangle inequality, x ∈ B(x0 , 1/n) ⊆ U , so by Lemma 13.2 we are done. Exercise 30.8. Which of our four countability axioms does Rω in the uniform topology satisfy? Solution. Rω is first countable since it is metrizable (see p. 130 and Example 30.2), but is not second countable by Example 30.2. By Exercise 30.5, we then see that Rω does not have a countable dense subset, and is also not Lindelöf. Exercise 30.9. Let A be a closed subspace of X. Show that if X is Lindelöf, then A is Lindelöf. Show by example that if X has a countable dense subset, A need not have a countable dense subset.

29

Proof. X is Lindelöf if and only if a collection of closed subsets of X with empty intersection has a countable subcollection with empty intersection by taking complements in Theorem 30.3(a). Now suppose we have a collection C of closed subsets of A with empty intersection; it is then also a collection of closed subsets of X with empty intersection by Theorem 17.3 since A is closed, and so has a countable subcollection with empty intersection since X is Lindelöf. Thus, A is also Lindelöf. Now let X = R2` . We see Q2 is countable, and is dense in X since if we take x ∈ X and a neighborhood U 3 x, there exists a basis element [a, b) × [c, d) ⊆ U containing x, which intersects Q2 by the fact that (a, b) × (c, d) ∩ Q2 6= ∅ since Q is dense in R. Thus, X has a countable dense subset; we claim that L = {x × (−x) | x ∈ R` } is a closed subspace of X that does not have a countable dense subset. L is closed since if (x1 , x2 ) ∈ X \ L, then letting d = x1 + x2 , the basis element [x1 − d/3, x1 + d/3), [x2 − d/3, x2 + d/3) does not intersect L. But then, L has the discrete topology since {(x, −x)} = L ∩ [x, b) × [−x, d) is open in L. Thus, if A ⊆ L, A = A by discreteness, and so A = L is true if and only if A = L. Thus, L has no countable dense subset. Exercise 30.17. Give Rω the box topology. Let Q∞ denote the subspace consisting of sequences of rationals that end in an infinite string of 0’s. Which of our four countability axioms does this space satisfy? Proof. We claim Q∞ is not first countable, and therefore not second countable. Suppose we have a countable basis {Ui } at 0 = (0, 0, 0, . . .) ∈ Qω . Let Vj ( πjQ (Uj ) open in Q with the subspace topology induced by R. Then, the neighborhood j Vj 3 0 does not contain any Ui , so {Ui } is not a basis and Q∞ is not first or second countable. We now show Q∞ has a countable dense subset. For, Qn = Qn × {0} × {0} are countable since are finite products of countable sets, and so their countable S they ∞ n union Q = Q is also countable. Thus, Q∞ is countable and so is a countable dense subset of itself. We now show Q∞ is Lindelöf. Suppose V is an open covering of Q∞ . Then, since ∞ Q is countable, choosing for every x ∈ Q∞ one element V ∈ V such that x ∈ V , we get a countable subcover of Q∞ .

31

The Separation Axioms

Exercise 31.3. Show that every order topology is regular. Proof. Let X be an ordered set with the order topology. X is Hausdorff and therefore T1 by Theorem 17.11. It therefore suffices to show the condition in Lemma 31.1(a). 30

So let x ∈ X and let U be an open neighborhood of x; we will construct a basis element V of the order topology such that x ∈ V and V ⊆ U . Suppose x is neither the smallest nor largest element of X. Then, x ∈ (a, b) ⊆ U for some basis element (a, b) of the order topology. If (a, x), (x, b) are nonempty, then let V = (u, v) where u ∈ (a, x) and v ∈ (x, b). If (a, x) = ∅, let u = a, so that V = (u, v) = [x, v); if (x, b) = ∅, let v = b, so that V = (u, v) = (u, x]. Then, x ∈ (u, v) and (u, v) ⊆ (a, b) ⊆ U . Now suppose x is the smallest (resp. largest) element of X. Then, x ∈ [x, b) ⊆ U (resp. x ∈ (a, x] ⊆ U ) for some basis element [x, b) (resp. (a, x]) of the order topology. Now if (x, b) (resp. (a, x)) is nonempty, then let V = [x, v) (resp. (u, x]) where v ∈ (x, b) (resp. u ∈ (a, x)). On the other hand, if (x, b) (resp. (a, x)) is empty, then let v = b (resp. u = a), so that V = {x}. We then have x ∈ V and V ⊆ U . If x is the smallest and largest element of X, then X = {x} is trivially regular.

32

Normal Spaces

Exercise 32.1. Show that a closed subspace of a normal space is normal. Proof. Suppose Y is our closed subspace of our normal space X, and A, B ⊆ Y disjoint and closed. By Theorem 17.3, A, B are closed in X. Let U, V be a separation of A, B in X. Then, Y ∩ U, Y ∩ V separate A, B in Y , and so Y is normal. Exercise 32.3. Show that every locally compact Hausdorff space is regular. Proof. Let X be a locally compact Hausdorff space; in particular it is T1 . Let x ∈ X with neighborhood U 3 x. By Theorem 29.2, there exists a neighborhood V 3 x such that V ⊆ U . By Lemma 31.1(a), X is then regular. Exercise 32.4. Show that every regular Lindelöf space is normal. Proof. Let A, B be disjoint closed subsets of X regular and Lindelöf. For all x ∈ A, there exists a neighborhood U 3 x disjoint from B. By regularity, there exists a neighborhood U ⊆ U ⊆ U containing x; since these U cover A, and A is Lindelöf by Exercise §30.9, there exists a countable subcover {Ui } such that Ui ∩ B = ∅ for all i. Similarly, we can construct a countable subcover {Vi } of B such that Vi ∩ A = ∅ for all i. By the exact same argument as in the proof of Theorem 32.1, then, the sets n n [ [ [ [ 0 0 U = Un \ Vi , V = Vn \ Ui n∈Z+

i=1

n∈Z+

i=1

are open and U 0 ⊇ A, V 0 ⊇ B, U 0 ∩ V 0 = ∅, and so X is normal. 31

Exercise 32.5. Is Rω normal in the product topology? In the uniform topology? Proof. Since Rω is metrizable in the product topology by Theorem 20.5 and in the uniform topology by definition on p. 124, both are normal by Theorem 32.2.

33

The Urysohn Lemma

Exercise 33.1. Examine the proof of the Urysohn lemma, and show that for given r, \ [ f −1 (r) = Up − Uq , p>r

q
p, q rational. Proof. ⊆. Suppose x ∈ f −1 (r), i.e., x ∈ Ur by definition, and x ∈ / Uq for all q < r by definition in Step 3. T Then, x ∈S Ur ⊆ Up for all p > r by construction in Steps 1, 2. ⊇. Suppose x ∈ p>r Up − q r, and so f (x) ≤ r by Step 4(1), and also x ∈ / Uq for all q < r, and so f (x) ≥ r by Step 4(2). Thus, f (x) = r.

34

The Urysohn Metrization Theorem

Exercise 34.3. Let X be a compact Hausdorff space. Show that X is metrizable if and only if X has a countable basis. Proof. ⇒. X is compact and metrizable, hence second countable by Exercise 30.4. ⇐. X is compact and Hausdorff, and so X is regular by Theorem 32.3. X is second countable as well, and so by the Urysohn metrization theorem (Theorem 34.1), X is metrizable. Exercise 34.5. Let X be a locally compact Hausdorff space. Let Y be the one-point compactification of X. Is it true that if X has a countable basis, then Y is metrizable? Is it true that if Y is metrizable, then X has a countable basis? Solution. If Y is metrizable, then it is second countable by Exercise 30.4. X is then second countable by Theorem 30.2. Now suppose S X has a countable basis B = {Bi }. We claim that B with sets of the form Y \ Bi , where the union is finite and the closure is taken in Y , form a basis of S Y ; call this larger basis B + . The Bi ∈ B are open by Lemma 16.2, and the Y \ Bi by Theorem 26.3. B + is countable since B is countable and there are S countably many Y \ Bi by Exercise 7.5(j). 32

Now recall that we have two types of open sets of Y , sets U open in X, and sets Y \ C for C compact in X, by construction in Theorem 29.1. By Lemma 13.2, it suffices to show that for each x in an open set, we can find an element of B + containing x properly contained in the open set. So first consider the U . For every x ∈ U , there exists Bi ∈ B such that x ∈ Bi ⊆ U since B is a basis of X. Now consider the Y \ C and x ∈ Y \ C. If x ∈ X, then Y \ C ∩ X 3 x is open in X by Theorem 29.1 (which says X is a subspace), and so the previous argument for U ⊆ X applies. It remains to show the case x = ∞, where {∞} = Y \ X. For each y ∈ C, we can find a neighborhood V 3 y in X such that V ⊆ X = Y \ {x} by local compactness. Since V is open in X, there exists some Bi ⊆ V containing y. S These B cover C and so there is a finite subcover by compactness C ⊆ B ; also, i S S i x∈ / Bi ⊇ C. Thus, we have x ∈ Y \ Bi ⊆ Y \ C. B + is therefore a basis for Y by Lemma 13.2. Since Y is compact and Hausdorff, it is normal by Theorem 32.3, and in particular, Y is regular. Finally, since Y also has a countable basis, it is metrizable by the Urysohn metrization theorem (Theorem 34.1).

36

Imbeddings of Manifolds

Exercise 36.1. Prove that every manifold is regular and hence metrizable. Where do you use the Hausdorff condition? Proof. Let X be our m-manifold. We first show X is locally compact. Let x ∈ X and a neighborhood U 3 x be given. Since X is a manifold, there exists a homeomorphism f : U → f (U ) ⊆ Rm . Since Rm is locally compact by Example 29.2, there exists a neighborhood V ⊆ f (U ) of f (x) such that V is compact and V ⊆ f (U ) by Theorem 29.2. Then, x ∈ f −1 (V ) ⊆ f −1 (V ) ⊆ U . But then, f −1 (V ) is compact and therefore closed by Theorems 26.3 and 26.5, and so f −1 (V ) ⊆ f −1 (V ) since the closure of a set is the intersection of all closed sets containing it, and moreover f −1 (V ) = f −1 (V ) by Theorem 18.1. Finally, we have x ∈ f −1 (V ) ⊆ f −1 (V ) ⊆ U , with f −1 (V ) compact, and so X is locally compact by Theorem 29.2. Now since X is locally compact and Hausdorff, X is regular by Exercise 32.3. Since X is regular and has a countable basis, it is metrizable by the Urysohn metrization theorem (Theorem 34.1). Note we used that X is Hausdorff in showing X is regular, for the characterization of local compactness, and the assertion that compact implies closed. Exercise 36.5. The Hausdorff condition is an essential part of the definition of a manifold; it is not implied by the other parts of the definition. Consider the following 33

space: Let X be the union of the set R − {0} and the two-point set {p, q}. Topologize X by taking as basis the collection of all open intervals in R that do not contain 0, along with all sets of the form (−a, 0) ∪ {p} ∪ (0, a) and all sets of the form (−a, 0) ∪ {q} ∪ (0, a), for a > 0. The space X is called the line with two origins. (a) Check that this is a basis for a topology. (b) Show that each of the spaces X − {p} and X − {q} is homeomorphic to R. (c) Show that X satisfies the T1 axiom, but is not Hausdorff. (d) Show that X satisfies all the conditions for a 1-manifold except for the Hausdorff condition. Proof of (a). For any x ∈ X, for a large enough either (−a, 0) ∪ {p} ∪ (0, a) or (−a, 0) ∪ {q} ∪ (0, a) contains x. Moreover, if we have two basis elements B1 , B2 , their intersection is either empty, already another basis element, or a set of the form (−a, 0) ∪ (0, a), we have a basis for X, for if x ∈ B1 ∩ B2 , B1 ∩ B2 , or in the last case choosing (−a, 0) or (0, a), would be a basis element containing x that is contained in the intersection. Proof of (b). Let f : X \ {q} → R be defined such that x 7→ 0 if x = p, and x 7→ x otherwise. Clearly f is a bijection; it suffices to show it is continuous and open. A basis element not containing p, q maps to a basis element not containing 0 by definition and vice versa. A basis element of the form (−a, 0) ∪ {p} ∪ (0, a) maps to (−a, a), and in the other direction, an open interval (−a, b) for a, b > 0 maps to (−a, 0) ∪ {p} ∪ (0, a) ∪ (0, b) if a < b, and similarly if a > b. Thus, f is a homeomorphism. Note g : X \ {p} → R is also a homemorphism by the same argument. Proof of (c). To show X is T1 , it suffices to show {x} is closed for all x ∈ X. {p}, {q} are closed since X \ {p} = (−∞, 0) ∪ {q} ∪ (0, ∞), X \ {q} = (−∞, 0) ∪ {p} ∪ (0, ∞) are open. {x} is closed for x 6= p, q since if x < 0, X \ {x} = (−∞, x) ∪ (x, 0) ∪ {p, q} ∪ (0, ∞) is open, and likewise for if x > 0. Note X is not Hausdorff since any neighborhood U of {p} intersects any neighborhood V of {q}, since U, V must contain basis elements Bp , Bq that contain p, q respectively; however, Bp ∩ Bq = ((−a, 0) ∪ {p} ∪ (0, a)) ∩ ((−b, 0) ∪ {p} ∪ (0, b)) = (−c, 0) ∪ (0, c) 6= ∅ for c = min{a, b}. Proof of (d). We claim the basis elements with rational end points form a countable basis. For any (a, b) basis element containing x, we can find a < r < s < b such that x ∈ (r, s) ⊆ (a, b); for any (−a, 0) ∪ {p} ∪ (0, a) basis element containing x, we can find 0 < r < a such that x ∈ (−r, 0) ∪ {p} ∪ (0, r) ⊆ (−a, 0) ∪ {p} ∪ (0, a). Thus they form a basis by Lemma 13.2. Now for any x 6= p, q, we see that there is a neighborhood U of x not containing p, q, and U is homeomorphic to a neighborhood in R by (b). For x = p, q, 34

any neighborhood of x will contain a basis element (−a, 0) ∪ {x} ∪ (0, a), which is homeomorphic to (−a, a) by (b).

Part II

Algebraic Topology 9

The Fundamental Group

51

Homotopy of Paths

Exercise 51.1. Show that if h, h0 : X → Y are homotopic and k, k 0 : Y → Z are homotopic, then k ◦ h and k 0 ◦ h0 are homotopic. Proof. Let F be the homotopy between h, h0 , and G the homotopy between k, k 0 . Let H : X × 1 → Z where H(x, t) = G(F (x, t), t). Then H(x, 0) = G(F (x, 0), 0) = G(f (x), 0) = (k ◦ h)(x), H(x, 1) = G(F (x, 1), 1) = G(f (x), 1) = (k 0 ◦ h0 )(x). It remains to show H is continuous. H is the map (x, t) 7→ (F (x, t), t) 7→ G(F (x, t), t); since G is already continuous and the composition of continuous functions is continuous, it suffices to show (x, t) 7→ (F (x, t), t) is continuous. But this is clear since this map is continuous in each coordinate in the codomain. Exercise 51.2. Given spaces X and Y , let [X, Y ] denote the set of homotopy classes of maps of X into Y . (a) Let I = [0, 1]. Show that for any X, the set [X, I] has a single element. (b) Show that if Y is path connected, the set [I, Y ] has a single element. Proof of (a). Fix f0 : X → I. For arbitrary f : X → I, f ' f0 by straight-line homotopy since any straight line in I is contained in I. Thus, since f was arbitrary, [X, I] = {[f0 ]}. Proof of (b). Let f : I → Y and y = f (0). Let g : I → Y be the constant map with value y ∈ Y . Define F : I × I → Y by F (s, t) = f (s(1 − t)). Since F (s, 0) = f (s) and F (s, 1) = g(s), we see F is a homotopy f ' g. Now fix y0 ∈ Y , and let ρ be a path connecting y, y0 . Define H : X × I → Y such that H(s, t) = ρ(t). Then, if f0 is the constant map with value y0 ∈ Y , H is a homotopy g ' f0 . By transitivity of homotopy, we see f ' f0 , and so since f was arbitrary, [X, Y ] = {[f0 ]}. Exercise 51.3. A space X is said to be contractible if the identity map iX : X → X is nulhomotopic. 35

(a) Show that I and R are contractible. (b) Show that a contractible space is path connected. (c) Show that if Y is contractible, then for any X, the set [X, Y ] has a single element. (d) Show that if X is contractible and Y is path connected, then [X, Y ] has a single element. Proof of (a). Since both I and R are convex, we see that any two maps are homotopic by straight-line homotopy as on pp. 324–325 since then the straight lines are fully contained in convex set. In particular, iX ' f for any constant map f . Proof of (b). If X is our contractible space, and F is our homotopy between iX and f0 for f0 a constant map with value x0 ∈ X, the map ρ : I → X where ρ(t) = F (x, t) is a path connecting x, x0 for any x ∈ X. Proof of (c). Let g0 : Y → Y be the constant map with value y0 ∈ Y ; we have iY ' g0 . Now define f0 : X → Y such that f0 (x) = y0 . Then, for any f : X → Y , we have f = iY ◦ f ' g0 ◦ f = f0 by Exercise 51.1, and so since f was arbitrary, [X, Y ] = {[f0 ]}. Proof of (d). Let g0 : X → X be the constant map with value x0 ∈ X; we have iX ' g0 . Now define y = f (x0 ). Then, for any f : X → Y , we have f = f ◦iX ' f ◦g0 , which is the constant map at y. Fix y0 ∈ Y with f0 : X → Y the constant map with value y0 ∈ Y , and let ρ be a path connecting y, y0 . Define H : X × I → Y such that H(x, t) = ρ(t). Then, H is a homotopy between f ◦ g0 and f0 . By transitivity of homotopy, we see f ' f0 , and so since f was arbitrary, [X, Y ] = {[f0 ]}.

52

The Fundamental Group

Exercise 52.4. Let A ⊂ X; suppose r : X → A is a continuous map such that r(a) = a for each a ∈ A. (The map r is called a retraction of X onto A.) If a0 ∈ A, show that r∗ : π1 (X, a0 ) −→ π1 (A, a0 ) is surjective. Proof. Letting ι : A ,→ X be the inclusion map, we see r◦ι = idA , and so by Theorem 52.4, r∗ ◦ ι∗ = (r ◦ ι)∗ = (idA )∗ = idπ1 (A,a0 ) . This implies r∗ is surjective.

36

53

Covering Spaces

Exercise 53.3. Let p : E → B be a covering map; let B be connected. Show that if p−1 (b0 ) has k elements for some b0 ∈ B, then p−1 (b) has k elements for every b ∈ B. In such a case, E is called a k-fold covering of B. Proof. Since p is a covering map, for any b with |p−1 (b)| = j there exists a neighborhood U 3 b such that p−1 (U ) is the disjoint union of j open neighborhoods Vα homeomorphic to U , for each Vα contains a unique preimage of b, which must be one point since U, Vα are homeomorphic. By the same argument, all x ∈ U are such that |p−1 (x)| = j. Thus, we can partition B into disjoint open sets Aj where each x ∈ Aj are such that p−1 (x) = j (note that j can be any cardinal since the argument above does not depend on j being finite), and Ak 3 b0 is one of the Aj ; however, since B is connected, B = Ak , for otherwise the Aj will be a separation of B. Exercise 53.5. Show that the map of Example 3 is a covering map. Generalize to the map p(z) = z n . Proof. Note that p(z) when considered as a map on S 1 as a subset of R2 is given by p(cos 2πx, sin 2πx) = (cos 2πnx, sin 2πnx). Letting q(x) = (2 cos πx, sin 2πx) be the covering R → S 1 in Theorem 53.1, and r : R → R be the multiplication by n, we then have the commutative diagram R2

r

q −1

R2 q

S1

p

S1

Now let x ∈ S 1 , and considerFU = S 1 \ {x}. Taking s ∈ q −1 (x), we see that we have a partition q −1 (U ) = i∈Z Wi where Wi = (s + i, s + i + 1), and we also have r−1 (W Fi ) = ((s + i)/n, (s + i + 1)/n) =: Zi . Letting Vi = q(Zi ), we see that −1 p (U ) = 0≤i
r|Wi

(q|Wi )−1

Vi

Zi q|Zi

p|Vi

U

commutes, where the top and vertical arrows are homeomorphisms, and so we have a homeomorphism between Vi and U for all i.

37

54

The Fundamental Group of the Circle

Exercise 54.1. What goes wrong with the “path-lifting lemma” (Lemma 54.1) for the local homeomorphism of Example 2 of §53? Solution. Since b0 = (1, 0) cannot be covered by an open set that is evenly covered by p by Example 53.2, we see that the first step in the proof of Lemma 54.1 fails. Exercise 54.4. Consider the covering map p : R × R+ → R2 − 0 of Example 6 of §53. Find liftings of the paths f (t) = (2 − t, 0), g(t) = ((1 + t) cos 2πt, (1 + t) sin 2πt), h(t) = f ∗ g. Sketch these paths and their liftings. Solution. We see first that the covering map p is the mapping (x, s) 7→ ((cos 2πx, sin 2πx), s) 7→ s(cos 2πx, sin 2πx). Thus, we have the family of liftings fñ (t) = (n, 2 − t) gñ (t) = (t + n, 1 + t) ( if t ∈ [0, 1/2] ˜ n (t) = (n, 2 − 2t) h (2t + n − 1, 2t) if t ∈ [1/2, 1] where n ∈ Z. We omit the sketches. Exercise 54.5. Consider the covering map p × p : R × R → S 1 × S 1 of Example 4 of §53. Consider the path f (t) = (cos 2πt, sin 2πt) × (cos 4πt, sin 4πt) in S 1 × S 1 . Sketch what f looks like when S 1 × S 1 is identified with the doughnut surface D. Find a lifting f˜ of f to R × R, and sketch it. Solution. We have the family of liftings fñ (t) = (n + t/2, n + t), where n ∈ Z. We omit the sketches. 38

58

Deformation Retracts and Homotopy Type

Exercise 58.1. Show that if A is a deformation retract of X, and B is a deformation retract of A, then B is a deformation retract of X. Proof. Let F : X × I → X, G : A × I → A be the deformation retractions of X onto A and A onto B respectively. We claim that ( F (x, 2t) if 0 ≤ t ≤ 1/2 H(x, t) = ι ◦ G(F (x, 1), 2t − 1) if 1/2 ≤ t ≤ 1 is a deformation retraction of X onto B, where ι : A ,→ X is the inclusion map. We see H(x, 0) = F (x, 0) = x, and that if x ∈ B, H(x, 1) = ι ◦ G(F (x, 1), 1) = ι ◦ F (x, 1) = ι ◦ x = x; moreover, H is continuous by the pasting lemma, since H(x, 1/2) = F (x, 1) = ι ◦ G(F (x, 1), 0) = ι ◦ F (x, 1) = F (x, 1), and since ι ◦ G(F (x, 1), 2t − 1) is a composition of continuous functions. Exercise 58.2. For each of the following spaces, the fundamental group is either trivial, infinite cyclic, or isomorphic to the fundamental group of the figure-eight. Determine for each space which of the three alternatives holds. (a) The “solid torus,” B 2 × S 1 . (b) The torus T with a point removed. (c) The cylinder S 1 × R. (d) The infinite cylinder S 1 × R. (e) R3 with the nonnegative x, y, and z axes deleted. The following subsets of R2 : (f ) {x | kxk > 1} (g) {x | kxk ≥ 1} (h) {x | kxk < 1} (i) S 1 ∪ (R+ × 0) (j) S 1 ∪ (R+ × R) (k) S 1 ∪ (R × 0) (l) R2 − (R+ × 0) Remark. We use Theorem 60.1 to say that π1 (−) and × commute. Solution. (a). π1 (B 2 × S 1 ) = π1 (B 2 ) × π1 (S 1 ) = π1 (S 1 ), for we have a deformation retraction from B 2 onto {0}. (b). π1 (T \ {p}) is isomorphic to the fundamental group of the figure-eight, for we have a deformation retraction onto the figure-eight by deforming T \ {p} in the following manner: 39

b a

b a

b

a

b a

a

b

a

a

b

b

where the last arrow comes from the construction of the torus as the quotient R2 /Z2 . (c). π1 (S 1 × I) = π1 (S 1 ) × π1 (I) = π1 (S 1 ), for we have the deformation retraction from I onto {0}. (d). π1 (S 1 × R) = π1 (S 1 ) × π1 (R), for we have the deformation retraction from R onto {0}. (e). This space retracts onto S 2 \ {p, q, r} ' R2 \ {s, t} (where the isomorphism is from Theorem 59.3), which retracts onto the figure-eight, and so the fundamental group is isomorphic to the fundamental group of the figure-eight. (f ). π1 ({x | kxk > 1}) = π1 (S 1 ), by a deformation retraction onto a circle of radius > 1, whose fundamental group is π1 (S 1 ) since it is homeomorphic to S 1 . (g). π1 ({x | kxk ≥ 1}) = πs (S 1 ), for we have the deformation retraction onto S 1 . (h). π1 ({x | kxk < 1}) = π1 ({0}) = 1, by a deformation retraction onto {0}. (i). π1 (S 1 ∪ (R+ × 0)) = π1 (S 1 ), since we can retract R+ to one point (1, 0) ∈ S 1 . (j). π1 (S 1 ∪ (R+ × R)) = π1 (S 1 ), since we can retract R+ × R to the half-circle 1 S ∩ R+ × R. (k). π1 (S 1 ∪ (R × 0)) is isomorphic to the fundamental group of the figure-eight, for we can retract (R × 0) onto the line segment from (−1, 0) to (1, 0), which gives a topological space homotopy equivalent to the figure-eight. (l). π1 (R2 \ (R+ × 0)) = 1, since we can retract onto any arbitrary point p ∈ 2 R \ (R+ × 0). Exercise 58.9. We define the degree of a continuous map h : S 1 → S 1 as follows: Let b0 be the point (1, 0) of S 1 ; choose a generator γ for the infinite cyclic group π1 (S 1 , b0 ). If x0 is any point of S 1 , choose a path α in S 1 from b0 to x0 , and define γ(x0 ) = α ˆ (γ). Then γ(x0 ) generates π(S 1 , x0 ). The element γ(x0 ) is independent of the choice of the path α, since the fundamental group of S 1 is abelian. Now given h : S 1 → S 1 , choose x0 ∈ S 1 and let h(x0 ) = x1 . Consider the homomorphism h∗ : π(S 1 , x0 ) −→ π1 (S 1 , x1 ). Since both groups are infinite cyclic, we have h∗ (γ(x0 )) = d · γ(x1 ) 40

(∗)

for some integer d, if the group is written additively. The integer d is called the degree of h and is denoted by deg h. The degree of h is independent of the choice of the generator γ; choosing the other generator would merely change the sign of both sides of (∗). (a) Show that d is independent of the choice of x0 . (b) Show that if h, k : S 1 → S 1 are homotopic, they have the same degree. (c) Show that deg(h ◦ k) = (deg h) · (deg k). (d) Compute the degrees of the constant map, the identity map, the reflection map ρ(x1 , x2 ) = (x1 , −x2 ), and the map h(z) = z n , where z is a complex number. (e) Show that if h, k : S 1 → S 1 have the same degree, they are homotopic. Proof of (a). Let x0 = 6 y0 ∈ S 1 , and let h(y0 ) = y1 . Let α be a path from x0 to y0 ; 0 then, β 0 = h ◦ β is a path from x1 to y1 . Then, γ(y1 ) = [β ] ∗ γ(x1 ) ∗ [β 0 ], and so 0

h∗ (γ(y0 )) = h∗ ([β] ∗ γ(x0 ) ∗ [β]) = [β ] ∗ h∗ (γ(x0 )) ∗ [β 0 ] 0

0

= [β ] ∗ d · γ(x1 ) ∗ [β 0 ] = d · ([β ] ∗ γ(x1 ) ∗ [β 0 ]) = d · γ(y1 ). Proof of (b). Choose x0 ∈ S 1 and let h(x0 ) = y0 , k(x0 ) = y1 . By Lemma 58.4, there exists a path α from y0 to y1 such that k∗ = α ˆ ◦ h∗ . Then, we have deg k · γ(y0 ) = k∗ (γ(x0 )) = α ˆ (h∗ (γ(x0 ))) =α ˆ (deg h · γ(y1 )) = deg h · α ˆ (γ(y1 )) = deg h · γ(y0 ), and so deg k = deg h. Proof of (c). Choose x0 ∈ S 1 and let k(x0 ) = y0 , h(y0 ) = y1 . Then, deg(h ◦ k) · γ(y1 ) = (h ◦ k)∗ (γ(x0 )) = (h∗ ◦ k∗ )(γ(x0 )) = h∗ (deg k · γ(y0 )) = deg k · h∗ (γ(y0 )) = (deg h) · (deg k)γ(y1 ), and so deg(h ◦ k) = (deg h) · (deg k). Solution for (d). Let k : S 1 → S 1 be the constant map. Then, we have k∗ (γ(x0 )) = 0, and so deg k = 0. Let id : S 1 → S 1 be the constant map. Then, we have id∗ (γ(x0 )) = γ(x0 ), and so deg id = 1. Let h : S 1 → S 1 such that z 7→ z n . Let p : R → S 1 be the standard covering. −1 p (b0 ) = 0, which is fixed by h. Let γ be the loop γ(s) = e2πis , which generates π1 (S1 , b0 ); note we can choose γ, b0 in this way by the fact that degree is independent of choice of generator and x0 . Then, (h ◦ γ)(s) = e2πins = p(ns), and so the lift of h ◦ γ is s 7→ ns, i.e., h∗ (γ(b0 )) = nγ(b0 ). Thus, deg h = n. The case for ρ follows from that of h, where n = −1. 41

Proof of (e). By (a), we use b0 as our base point, and let h(b0 ) = x0 , k(b0 ) = y0 . Consider h∗ : π1 (S 1 , b0 ) → π1 (S 1 , x0 ), k∗ : π1 (S 1 , b0 ) → π1 (S 1 , y0 ); by assumption, there exists n, such that h∗ (γ(b0 )) = n · γ(x0 ), k∗ (γ(b0 )) = n · γ(y0 ). Now let γ(s) = e2πis , which generates π1 (S1 , b0 ) as above. Then, let h] ◦ γ, k] ◦ γ be 1 the lifts of h ◦ γ, k ◦ γ through the standard covering p : R → S , at h0 ∈ p−1 (x0 ), k0 ∈ p−1 (y0 ) respectively. Let h1 = h] ◦ γ(1), k1 = k] ◦ γ(1). Now we have h∗ (γ(b0 )) = (h1 − h0 )γ(x0 ), k∗ (γ(b0 )) = (k1 − k0 )γ(y0 ), and so h1 − h0 = k1 − k0 = n. We want to find a path homotopy from h] ◦ γ to k] ◦ γ. We ] ] first define h] ◦ γ = h] ◦ γ − h0 + k0 so that both h] ◦ γ, h] ◦ γ have the same end points; ˜ we then see there exists a path homotopy F between them since R is contractible. ] ˜ ˜ ◦ γ at k0 . Then, Now consider h(x) = e2πi(k0 −h0 ) h(x). Then, h] ◦ γ is the lift of h ] ◦ γ to k] ◦ γ. Since this is a path homotopy, it factors p◦ F˜ is a path homotopy from h] ˜ to k. through (p, id) to get a homotopy F from h ˜ and h given by We now see that there exists a path homotopy G between h 2πis(k0 −h0 ) G(x, t) = h(x)e , and so there exists a path homotopy combining F, G between h and k. Exercise 58.10. Suppose that to every map h : S n → S n we have assigned an integer, denoted by deg h and called the degree of h, such that (i) Homotopic maps have the same degree. (ii) deg(h ◦ k) = (deg h) · (deg k). (iii) The identity has degree 1, any constant map has degree 0, and the reflection map ρ(x1 , . . . , xn+1 ) = (x1 , . . . , xn , −xn+1 ) has degree −1. Prove the following: (a) There is no retraction r : B n+1 → S n . (b) If h : S n → S n has degree different from (−1)n+1 , then h has a fixed point. (c) If h : S n → S n has degree different from 1, then h maps some point x to is antipode −x. (d) If S n has a nonvanishing tangent vector field v, then n is odd. Proof of (a). We use polar coordinates to denote B n+1 and S n , i.e., the points in B n+1 are given by (θ, φ2 , . . . , φn , ρ) and the points in S n by (θ, φ2 , . . . , φn ), where 0 ≤ θ < 2π, 0 ≤ φi < π, and 0 ≤ ρ ≤ 1. Thus, the inclusion S n ,→ B n+1 is given by (θ, φ2 , . . . , φn ) 7→ (θ, φ2 , . . . , φn , 1). Now, suppose a retraction r : B n+1 → S n exists. Then, define the homotopy H : S n × [0, 1] → S n by H(x1 , . . . , xn , t) = r(x1 , . . . , xn , t), which is continuous since r is. Then, H(x1 , . . . , xn , 0) is constant, hence has degree 0, whereas H(x1 , . . . , xn , 1) is the identity, hence has degree 1, which contradicts (i). 42

Proof of (b). We return to Cartesian coordinates. Suppose h has no fixed points. We first construct a homotopy: H(x, t) =

(1 − t)h(x) + ta(x) . k(1 − t)h(x) + ta(x)k

Since H is continuous since it is a composition of continuous functions in both variables, it suffices to show k(1 − t)h(x) + ta(x)k = 6 0 for all x, t. So, suppose (1 − t)h(x) + ta(x) = 0. Then, (1 − t)h(x) = tx. Comparing norms, this is only possible when t = 1/2, and so h(x) = x, which contradicts that h has no fixed points. Thus, we see that h is homotopic to the antipodal map. Now, since the antipodal map is the composition of n + 1 reflections, each one reflecting each coordinate of S n , we have that deg(h) = deg(a) = (−1)n+1 by (i), (ii), and (iii), a contradiction. Proof of (c). Suppose h maps no point x to its antipode −x. We first construct a homotopy: (1 − t)h(x) + tx . G(x, t) = k(1 − t)h(x) + txk Since G is continuous since it is a composition of continuous functions in both variables, it suffices to show k(1 − t)h(x) + txk 6= 0 for all x, t. So, suppose (1 − t)h(x) + tx = 0. Then, (1 − t)h(x) = −tx. Comparing norms, this is only possible when t = 1/2, and so h(x) = −x, which contradicts that h maps no point x to its antipode −x. Thus, we see that h is homotopic to the identity. Since the identity has degree 1, we have deg(h) = 1 by (i), a contradiction. Proof of (d). Suppose such a v(x) exists. Then, let h(x) = v(x)/kv(x)k; h(x) then is a map S n → S n . Since v(x) is a tangent vector field, hh(x), xi = hv(x), xi /kv(x)k = 0, where h·, ·i is the standard inner product in Rn+1 . Then, there are no points h(x) = ±x since hh(x), xi = 0 6= ±1. By the proofs of (b) and (c), h(x) is then homotopic to the identity and the antipodal map; hence, the identity is homotopic to the antipodal map. By (i), (iii), and the proofs of (b) and (c), this then implies that deg 1 = deg(−1)n+1 , and so n is odd.

59

The Fundamental Group of S n

Exercise 59.1. Let X be the union of two copies of S 2 having a single point in common. What is the fundamental group of X? Prove that your answer is correct.

43

Solution. Let U = X \ {a}, V = X \ {b} such that a, b are in different copies of S 2 ; U, V are open since X is Hausdorff. Then, since S 2 \ {a}, S 2 \ {b} are homeomorphic to R2 by the proof in Theorem 59.3, we see that there exists a deformation retraction from X \ {a} onto S 2 by taking the copy of S 2 containing a and retracting it into {x0 } the intersection of the two copies of S 2 ; likewise, there exists a deformation retraction from X \ {b} onto S 2 . Thus, both U, V are simply connected by Theorem 59.3. U ∩ V is path connected since it is homeomorphic to two copies of R2 adjoined at x0 , and so X is simply connected by Corollary 59.2.

60

Fundamental Groups of Some Surfaces

Exercise 60.2. Let X be the quotient space obtained from B 2 by identifying each point x of S 1 with its antipode −x. Show that X is homeomorphic to the projective plane P 2 . Proof. Consider X as constructed from B 2 identified with the closed upper hemisphere of S 2 . Let p : S 2 → P 2 , q : B 2 → X be the quotient maps, and π : S 2 → B 2 the map sending x to either itself or −x if x ∈ / B 2 ; note this is a quotient map since 2 2 U ⊆ S open implies π(U ) open, and V ⊆ B open implies π −1 (V ) = V ∪ −V open. We then have the commutative diagram S2

π

p

P2

B2 q

f

X

and since q ◦ π is a quotient map by p. 141, we see f is a quotient map by Theorem 22.2. But since for x ∈ X, (q◦π)−1 (x) = {x, −x} ∈ P 2 , and moreover any equivalence class {x, −x} can be realized as the inverse image of this type, f is a bijection and P 2 = {(q ◦ π)−1 (x) | x ∈ X}, and so f is a homeomorphism by Corollary 22.3(a). Exercise 60.3. Let p : E → X be the map constructed in the proof of Lemma 60.5. Let E 0 be the subspace of E that is the union of the x-axis and the y-axis. Show that p|E 0 is not a covering map. Proof. Consider the base point x0 , the center of the figure-eight X. A neighborhood U 3 x0 contains the union V of open intervals in A and B which intersect in exactly {x0 }. (p|E 0 )−1 (V ) is then equal to the union of open intervals around integers on the x-axis and open intervals around integers on the y-axis. But none of these intervals are homeomorphic to V , since removing an integer in an interval gives two connected components, while removing its image x0 in V gives four connected components. 44

11 67

The Seifert-van Kampen Theorem Direct Sums of Abelian Groups

Exercise 67.2. Show that if G1 is a subgroup of G, there may be no subgroup G2 of G such that G = G1 ⊕ G2 . Proof. Let G = Z and G1 = 2Z, and suppose that such a G2 exists. Then, we have G = 2Z ⊕ Z/2Z by Corollary 67.3, and so (0, 1) ∈ G1 ⊕ G2 . But since 2(0, 1) = (0, 0), G1 ⊕ G2 contains an element of order 2 while G does not, a contradiction. Exercise 67.4. The order of an element a of an abelian group G is the smallest positive integer m such that ma = 0, if such exists; otherwise, the order of a is said to be infinite. The order of a thus equals the order of the subgroup generated by a. (a) Show the elements of finite order in G form a subgroup of G, called its torsion subgroup. (b) Show that if G is free abelian, it has no elements of finite order. (c) Show the additive group of rationals has no elements of finite order, but is not free abelian. Proof of (a). It suffices to show the torsion elements T (G) of G is closed under sums and inverses. If a, b ∈ T (G) with orders m, n respectively, then mn(a + b) = n(ma) + m(nb) = 0 =⇒ a + b ∈ T (G); likewise, m(−a) = −(ma) = 0 =⇒ −a ∈ T (G). P Proof of (b). Any 0 6= a ∈ G is of the form a = kα aα where {aα } is our basis of G and kα ∈ Z. If ma = 0, then mkα aα = 0 for all α, which is a contradiction since each aα also has infinite order. Proof of (c). Suppose Q = haα i, {aα } a basis. Then P for any aα , aα /2 must be of the P form aα /2 = kα aα for kα ∈ Z. But then, aα = 2 kα aα , and since aα is a basis element, we have aα = 2kα aα . This implies kα = 1/2, which is a contradiction.

68

Free Products of Groups

Exercise 68.2. Let G = G1 ∗ G2 , where G1 and G2 are nontrivial groups. (a) Show G is not abelian. (b) If x ∈ G, define the length of x to be the length of the unique reduced word in the elements of G1 and G2 that represents x. Show that if x has even length (at least 2), then x does not have finite order. Show that if x has odd length (at least 3), then x is conjugate to an element of shorter length. 45

(c) Show that the only elements of G that have finite order are the elements of G1 and G2 that have finite order, and their conjugates. Proof of (a). For 1 6= g ∈ G1 , 1 6= h ∈ G2 , gh 6= hg since otherwise we would have distinct reduced word representations of the same element of G. Proof of (b). If x ∈ G has even length, then without loss of generality, x starts with an element in G1 and ends with one in G2 , and so we cannot reduce xn to the identity, and x has infinite order. If x ∈ G has odd length, then without loss of generality, x = ghg 0 for g, g 0 ∈ G1 , h ∈ G. Then, g −1 xg = hg 0 g has shorter length, for g 0 g reduces to one element in G1 . Proof of (c). Suppose x ∈ G has finite order. Then, by (b) it must have odd length 2k + 1. We proceed by induction on k. For k = 0, we see the length is 1, and so x ∈ Gi for some i, and has finite order in Gi . Now suppose k > 0. Since x has odd length, x = g −1 yg for g, y ∈ G, y of shorter length. y has odd length for if not, y has infinite order by (b), and so x also has infinite order since xn = g −1 y n g, which is a contradiction. Since y is of finite order by the fact gxn g −1 = y n , y is either equal to an element of Gi with finite order or conjugate to one by inductive hypothesis. If the latter is true, x = g −1 yg = g −1 h−1 zhg = (hg)−1 zhg for h ∈ G, and z ∈ Gi having finite order for some i. x is therefore conjugate to a finite order element of Gi .

71

The Fundamental Group of a Wedge of Circles

Exercise 71.2. Suppose X is a space that is the union of the closed subspaces X1 , . . . , Xn ; assume there is a point p of X such that Xi ∩ Xj = {p} for i 6= j. Then we call X the wedge of the spaces X1 , . . . , Xn , and write X = X1 ∨ · · · ∨ Xn . Show that if for each i, the point p is a deformation retract of an open set Wi of Xi , then π1 (X, p) is the external free product of the groups π1 (Xi , p) relative to the monomorphisms induced by inclusion. Proof. By induction, it suffices to show when X = X1 ∨ X2 ; moreover, it suffices to consider when the Xi are both path-connected since if Ci are the path components containing p in Xi , then π1 (Ci , p) = π1 (Xi , p) as on p. 332. So, let U = X1 ∪ W2 and V = X2 ∪ W1 . Then, both U and V are path connected since they deformation retract to X1 , X2 , respectively, and U ∩ V = W1 ∪ W2 is moreover simply connected since it deformation retracts to {p}. Thus, by Corollary 70.3, there is an isomorphism π1 (X1 , p) ∗ π1 (X2 , p) ' π1 (X, p). Exercise 71.4. Show that if X is an infinite wedge of circles, then X does not satisfy the first countability axiom. 46

Proof. Let p ∈ X be the common point of the circles. Suppose X has a countable basis {Ui } at p; we can assume without loss of generality that Ui ) Uj if i < j. For each Ui , we know that Vij = Ui ∩ Sj is open for any i, j by the coherence condition, and is nonempty since S p ∈ Vij . Now take V = Vii ; we claim it is a neighborhood of p that does not contain any Ui . It is open by coherence since V ∩ Si = Ui is open for all i, and contains p by construction above. Now suppose Ui ⊆ V . Then, this implies Ui ∩ Sj ⊆ V ∩ Sj = Uj for all j, and in particular when i < j, which contradicts that Ui ) Uj if i < j. Exercise 71.5. Let Sn be the circle of radius n in R2 whose center is at the point (n, 0). Let Y be the subspace of R2 that is the union of these circles; let p be their common point. (a) Show that Y is not homeomorphic to a countably infinite wedge X of circles, nor to the space of Example 1. (b) Show, however, that π1 (Y, p) is a free group with {[fn ]} as a system of free generators, where fn is a loop representing a generator of π1 (Sn , p). Proof of (a). Y is a subspace of R2 and so it is first countable by Theorem 30.2, and so Y is not homeomorphic to X by Exercise 71.4, which says X is not first countable. Denote Z as the space of Example 71.1; we claim Z is compact while Y is not. First, denoting Dm as the closed disc of radius 1/m with center at the point (1/m, 0), we seeSthat Z ∪ Dm is closed T since it is a finite union of closed sets Z ∪ Dm = Dm ∪ n
∗ ∞

n=1

π1 (Sn , p) → π1 (Y, p),

J J Y Y `j [fmj ] 7→ in ([fmj ])`j j=1

(1)

j=1

is an isomorphism. We first show that the homomorphism (1) is surjective. So, suppose f : I → Y S∞ is a loop in Y . Letting YN = Y \ n=N +1 {(2n, 0)}, since I is compact, the image f (I) is also compact, S thus bounded by Theorem 27.3, and so f (I) ⊆ YN for some N . Letting rN : Y → N n=1 Sn , we form the deformation retraction H : YN × I → YN , where H(x, 0) = idYN and H(x, 1) = rN |YN , by retracting the upper and lower 47

semicircles of Sn to p for all n > N . Then, H ◦ (f × idI ) : I × I → YN is a path homotopy from f to rN ◦ f . Thus, [f ] = [rN ◦ f ], and Theorem 71.1 implies that [rN ◦ f ] is a product of elements of the groups Gn for n ≤ N . Therefore, [f ] is in the image of the homomorphism (1). We now show that the homomorphism (1) is injective. Now suppose there is some non-identity element

∗ ∞

w∈

π1 (Sn , p),

n=1

whose image through the homomorphism (1) is the identity in π1 (Y, p). Let f be a loop in X whose path-homotopy class is the image of w. Then, f is path homotopic to a constant in X, so by the argument above, it is path homotopic to a constant in some S YN , and therefore path homotopic to a constant in N S . n=1 n But this contradicts Theorem 71.1, which says that the map ! N N _ π1 (Sn , p) → π1 Sn , p

∗

n=1

n=1

is injective.

73

The Fundamental Groups of the Torus and the Dunce Cap

Exercise 73.1. Find spaces whose fundamental groups are isomorphic to the following groups. (Here Z/n denotes the additive group of integers modulo n.) (a) Z/n × Z/m. (b) Z/n1 × Z/n2 × · · · × Z/nk . (c) Z/n ∗ Z/m. (See Exercise 2 of §71). (d) Z/n1 ∗ Z/n2 ∗ · · · ∗ Z/nk . Solution. We use Theorem 60.1 to say π1 (X × Y, x0 × y0 ) = π1 (X, x0 ) × π1 (Y, y0 ), and use Exercise 71.2 to say π1 (X ∨ Y, x0 ) = π1 (X, x0 ) ∗ π1 (Y, x0 ) for {x0 } = X ∩ Y . We denote the n-fold dunce cap by Dn . Note since the Dn are path connected, we do not need to specify base points. Through repeated applications of the above, and by using induction to get (a) ⇒

48

(b), (c) ⇒ (d), we get π1 (Dn × Dm ) = Z/n × Z/m =⇒ π1 π1 (Dn ∨ Dm ) = Z/n ∗ Z/m =⇒ π1

k Y i=1 k _

! Dni =

Classification of Surfaces

74

Fundamental Groups of Surfaces

Z/ni ,

i=1

! Dni

i=1

12

k Y

∗ k

=

Z/ni .

i=1

Exercise 74.3. The Klein bottle K is the space obtained form a square by means of the labeling scheme aba−1 b. Figure 74.11 indicates how K can be pictured as an immersed surface in R3 . (a) Find a presentation for the fundamental group of K. (b) Find a double covering map p : T → K, where T is the torus. Describe the induced homomorphism of fundamental groups. Proof of (a). π1 (K) = ha, b | aba−1 b = 1i by Theorem 74.2. Proof of (b). We consider T as [0, 1] × [0, 1] with the relations (0, y) ∼ (1, y), (x, 0) ∼ (x, 1), and K as [0, 1] × [0, 1] with the relations (0, y) ∼ (1, 1 − y), (x, 0) ∼ (x, 1). Then, define p : T → K by ( (2x, y) if x ∈ [0, 1/2], p(x, y) = (2x − 1, 1 − y) if x ∈ [1/2, 1]. This is continuous in each region, and agrees on the boundary since p(1/2, y) = (2 · 1/2, y) = (1, y) = (0, 1 − y) = (2 · 1/2 − 1, 1 − y) and p(1, 1 − y) = (1, 1 − y) = (0, y) = p(0, y). Now recall that π1 (T ) = hα, β | αβα−1 β −1 = 1i. Looking at Figures 74.4 and 74.11, we see p∗ (α) = a2 , p∗ (β) = b. Since aba−1 b = 1 ⇐⇒ bab = a, we see that p∗ (α)p∗ (β) = a2 b = babab = ba2 = p∗ (β)p∗ (α).

75

Homology of Surfaces

Exercise 75.3. Let X be the quotient space obtained from an 8-sided polygonal region P by pasting its edges together according to the labelling scheme acadbcb−1 d. 49

(a) Check that all vertices of P are mapped to the same point of the quotient space X by the pasting map. (b) Calculate H1 (X). (c) Assuming X is homeomorphic to one of the surfaces given in Theorem 75.5 (which it is), which surface is it? a c

d b

a c

d b−1

Figure 1: Labeling of edges and identification of vertices in P . Proof of (a). We have the identification of vertices in P as shown with solid lines in Figure 1, found by identifying heads/tails of arrows with the same label. Thus, all the vertices of P are mapped to the same point of the quotient space X by the pasting map. Solution for (b). We apply Lemma 77.1 to the labeling scheme acadbcb−1 d repeatedly to find an equivalent labeling scheme, where the brackets show our decomposition [y0 ]a[y1 ]a[y2 ]: [ ]a[c]a[dbcb−1 d] ∼ aac−1 dbcb−1 d [aac−1 ]d[bcb−1 ]d[ ] ∼ ddaac−1 bc−1 b−1 [ddaa]c−1 [b]c−1 [b−1 ] ∼ c−1 c−1 ddaab−1 b−1 . Relabeling the edges, which is allowed by p. 460, we have the labeling scheme aabbccdd. Thus, by Theorem 74.2, the fundamental group is π1 (X, x0 ) = ha, b, c, d | a2 b2 c2 d2 = 1i . Finally, the first homology group is H1 (X) ' ha, b, c, d | 2a + 2b + 2c + 2d = 0i ' Z3 × (Z/2Z).

50

Solution for (c). Theorem 75.4 says that H1 (Pn ) = Zn−1 × (Z/2Z). Thus, assuming that X is homeomorphic to one of the surfaces in Theorem 75.5, we see that X is homeomorphic to P4 = P 2 #P 2 #P 2 #P 2 by (b). Exercise 75.4. Let X be the quotient space obtained from an 8-sided polygonal region P by means of the labelling scheme abcdad−1 cb−1 . Let π : P → X be the quotient map. (a) Show that π does not map all the vertices of P to the same point of X. (b) Determine the space A = π(Bd P ) and calculate its fundamental group. (c) Calculate π1 (X, x0 ) and H1 (X). (d) Assuming X is homeomorphic to one of the surfaces given in Theorem 75.5, which surface is it? c

d

b

a

a b−1

d−1 c

Figure 2: Labeling of edges and identification of vertices in P . Proof of (a). We have the identification of vertices in P as shown in Figure 2, found by identifying heads/tails of arrows with the same label, where the solid lines are one identification and the dashed lines are another. Thus, π does not map all the vertices of P to the same point of X. b a

x0

x1

c

d Figure 3: Sketch of A = π(Bd P ). Solution for (b). Let x0 be the point identified by the solid lines in Figure 2, and x1 the point identified by the dashed lines. Then, using Figure 2, we see that a connects 51

x0 to itself, c connects x1 to itself, and b goes from x0 to x1 , while d goes from x1 to x0 . Thus, we have the sketch in Figure 3 for A = π(Bd P ). We now want to calculate its fundamental group. First, we see that we can homotopically retract the segment d into the point x0 , thereby making x0 coincide with x1 ; the resulting deformation retract is then the wedge sum of three circles. Thus, π1 (A, x0 ) ' π1 (S 1 ∨ S 1 ∨ S 1 , x0 ) = Z ∗ Z ∗ Z by Theorems 58.3 and 71.1. Solution for (c). We proceed as in Exercise 75.3(b): [ ]a[bcd]a[d−1 cb−1 ] ∼ aad−1 c−1 b−1 d−1 cb−1 [aad−1 c−1 ]b−1 [d−1 c]b−1 [ ] ∼ b−1 b−1 aad−1 c−1 c−1 d [b−1 b−1 aad−1 ]c−1 [ ]c−1 [d] ∼ c−1 c−1 b−1 b−1 aad−1 d c−1 c−1 b−1 b−1 aad−1 d ∼ c−1 c−1 b−1 b−1 aa c−1 c−1 b−1 b−1 aa ∼ aabbcc where we cancel d−1 d in the penultimate step and relabel in the last step as allowed on p. 460. Thus, our fundamental group is π1 (X, x0 ) = ha, b, c | a2 b2 c2 = 1i , which has the first homology group H1 (X) = ha, b, c, d | 2a + 2b + 2c = 0i ' Z2 × (Z/2Z). Solution for (d). Since H1 (X) ' Z2 × (Z/2Z), by Theorem 75.4, we have that X is homeomorphic to P3 = P 2 #P 2 #P 2 , assuming that X is homeomorphic to one of the surfaces in Theorem 75.5.

52

List of Solved Exercises I

General Topology

3 24

1

Set Theory and Logic 7 Countable and Uncountable Sets Exercise 7.5 . . . . . . . . . . . .

2

Topological Spaces and Continuous Functions 13 Basis for a Topology . . . . . . . Exercise 13.3 . . . . . . . . . . . Exercise 13.5 . . . . . . . . . . . Exercise 13.6 . . . . . . . . . . . Exercise 13.7 . . . . . . . . . . . 16 The Subspace Topology . . . . . Exercise 16.8 . . . . . . . . . . . Exercise 16.9 . . . . . . . . . . . 17 Closed Sets and Limit Points . . Exercise 17.2 . . . . . . . . . . . Exercise 17.3 . . . . . . . . . . . Exercise 17.5 . . . . . . . . . . . Exercise 17.13 . . . . . . . . . . . Exercise 17.16 . . . . . . . . . . . 18 Continuous Functions . . . . . . . Exercise 18.1 . . . . . . . . . . . Exercise 18.12 . . . . . . . . . . . 19 The Product Topology . . . . . . Exercise 19.6 . . . . . . . . . . . 20 The Metric Topology . . . . . . . Exercise 20.4 . . . . . . . . . . . Exercise 20.5 . . . . . . . . . . . Exercise 20.6 . . . . . . . . . . . Exercise 20.8 . . . . . . . . . . . 21 The Metric Topology (continued) Exercise 21.1 . . . . . . . . . . . Exercise 21.2 . . . . . . . . . . . Exercise 21.3 . . . . . . . . . . . 22 The Quotient Topology . . . . . . Exercise 22.2 . . . . . . . . . . . Exercise 22.4 . . . . . . . . . . . Exercise 22.6 . . . . . . . . . . .

3

3 3 3 25 3 3 3 4 4 4 5 5 6 6 4 6 6 7 7 7 8 8 8 9 9 10 10 12 12 13 15 15 16 16 18 18 18 19

Connectedness and Compactness 20 23 Connected Spaces . . . . . . . . . 20 Exercise 23.8 . . . . . . . . . . . 20

II

53

27

29

Exercise 23.11 . . . . . . . . . . Connected Subspaces of the Real Line . . . . . . . . . . . . Exercise 24.7 . . . . . . . . . . Exercise 24.8 . . . . . . . . . . Exercise 24.12 . . . . . . . . . . Components and Local Connectedness . . . . . . . . . . . . Exercise 25.2 . . . . . . . . . . Compact Subspaces of the Real Line . . . . . . . . . . . . . . . Exercise 27.4 . . . . . . . . . . Local Compactness . . . . . . . Exercise 29.4 . . . . . . . . . . Exercise 29.8 . . . . . . . . . .

Countability and Separation Axioms 30 The Countability Axioms . . . Exercise 30.4 . . . . . . . . . . Exercise 30.5 . . . . . . . . . . Exercise 30.8 . . . . . . . . . . Exercise 30.9 . . . . . . . . . . Exercise 30.17 . . . . . . . . . . 31 The Separation Axioms . . . . Exercise 31.3 . . . . . . . . . . 32 Normal Spaces . . . . . . . . . Exercise 32.1 . . . . . . . . . . Exercise 32.3 . . . . . . . . . . Exercise 32.4 . . . . . . . . . . Exercise 32.5 . . . . . . . . . . 33 The Urysohn Lemma . . . . . . Exercise 33.1 . . . . . . . . . . 34 The Urysohn Metrization Theorem . . . . . . . . . . . . . . . . Exercise 34.3 . . . . . . . . . . Exercise 34.5 . . . . . . . . . . 36 Imbeddings of Manifolds . . . . Exercise 36.1 . . . . . . . . . . Exercise 36.5 . . . . . . . . . .

Algebraic Topology

. 20 . . . .

21 21 22 22

. 26 . 26 . . . . .

27 27 27 27 28

. . . . . . . . . . . . . . .

28 28 28 29 29 29 30 30 30 31 31 31 31 32 32 32

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32 32 32 33 33 33

35

9

The Fundamental Group 51 Homotopy of Paths . . . . . . . Exercise 51.1 . . . . . . . . . . Exercise 51.2 . . . . . . . . . . Exercise 51.3 . . . . . . . . . . 52 The Fundamental Group . . . . Exercise 52.4 . . . . . . . . . . 53 Covering Spaces . . . . . . . . . Exercise 53.3 . . . . . . . . . . Exercise 53.5 . . . . . . . . . . 54 The Fundamental Group of the Circle . . . . . . . . . . . . . . Exercise 54.1 . . . . . . . . . . Exercise 54.4 . . . . . . . . . . Exercise 54.5 . . . . . . . . . . 58 Deformation Retracts and Homotopy Type . . . . . . . . . . Exercise 58.1 . . . . . . . . . . Exercise 58.2 . . . . . . . . . . Exercise 58.9 . . . . . . . . . . Exercise 58.10 . . . . . . . . . . 59 The Fundamental Group of S n Exercise 59.1 . . . . . . . . . . 60 Fundamental Groups of Some Surfaces . . . . . . . . . . . . .

. . . . . . . . .

35 35 35 35 11 The 35 67 36 36 37 68 37 37 71

. . . .

38 38 38 38

. . . . . . .

73

Exercise 60.2 . . . . . . . . . . . 44 Exercise 60.3 . . . . . . . . . . . 44 Seifert-van Kampen Theorem Direct Sums of Abelian Groups . Exercise 67.2 . . . . . . . . . . . Exercise 67.4 . . . . . . . . . . . Free Products of Groups . . . . . Exercise 68.2 . . . . . . . . . . . The Fundamental Group of a Wedge of Circles . . . . . . . . . Exercise 71.2 . . . . . . . . . . . Exercise 71.4 . . . . . . . . . . . Exercise 71.5 . . . . . . . . . . . The Fundamental Groups of the Torus and the Dunce Cap . . . . Exercise 73.1 . . . . . . . . . . .

39 39 39 40 12 Classification of Surfaces 74 Fundamental Groups of Surfaces 42 Exercise 74.3 . . . . . . . . . . . 43 75 Homology of Surfaces . . . . . . . 43 Exercise 75.3 . . . . . . . . . . . . 44 Exercise 75.4 . . . . . . . . . . .

54

45 45 45 45 45 45 46 46 46 47 48 48 49 49 49 49 49 51

284185717-Munkres-Solucionario.pdf

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