Sean Li Math 4530 − Cornell University Homework #9 Munkres,
Section 34, #4.
Let X be a locally compact compact Hausdorff Hausdorff space. Then it is completely completely regular from Exercise Exercise 33.7 (this can be shown by taking a one-point compactification of the space, which is normal, which which is completely completely regular). regular). By the Urysohn Urysohn metrization metrization theorem, regular and countable countable basis imply that X is metrizable, so the first statement is true. The second statement is false, as an uncountable set in the discrete topology is not first countable, yet some are metrizable (e.g., R with the discrete discrete topology). Section 35, #4.
(a) Use Exercise 31.5, which says that if f, g : X → W are continuous and W is Hausdorff, Hausdorff, then the set {x|f ( case,, let f : Z → Z be the retract f (x) = g(x)} is closed in X . In our case map r taking values in Z , an extended space of Y of Y ,, and let g : Z → Z be the identity map. Z is given to be Hausdorff, f is given to be continuous, and g is automatically continous as it is the ident identit ity y map. map. Then Then the cited cited exercis exercisee gives gives that the set {x|r(x) = x} is closed, showing that Y is closed. (b) Suppose a retract r from R2 to the two-point {a, b} set set exist exists. s. Let Let A = r−1 (a) and Now a and b are both clopen in the set {a, b}. Sinc Sincee r is continous, both B = r−1 (b). Now A and B are open. Moreover, they are disjoint as a point cannot map to both a and b, and they are nonempty as a ∈ A and b ∈ B . Th Then en A and B form a separation of R2 , contradicting R2 being a connected space.
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(c) The continous map r : (x, y) → (x, y )/ x + y 2 satisfies the retraction S 1 of R2 − {0}. It does not seem possible for S 1 to be a retract of R2 , as R2 − { 0} has a “hole” at 0 but R2 does not have that hole. The unit circle can be homotoped to a point in R2 but not in R2 − {0}. Anothe Anotherr way to think think of this this is that that S 1 is homeomorphic to the solid torus, and so is R2 − {0} when considered in the one-point compactification R2 ∪ {∞}. However, R2 is not homeomorphic to the solid torus. Section 35, #5.
(a) By the Tietze extension theorem, as X is normal and A ⊂ X is closed, each f j : A → R can be extended to g j : X → R continous. Thus f = (f j ) can be extended as g = (g j ), g : X → RJ . (b) Let f : A → Y be continous as stated in the problem. Let g : Y → B be a homeomorphism to a retract B of RJ , and let r : RJ → B be the retraction giving this retract. Then by part (a), there exists h : X → RJ which extends g ◦ f . f . Then the universal extension property is met as there is a continous extension of f given by g −1 ◦ r ◦ h : X → Y . Y .
Section 35, #6.
(a) Let g : Y → Y 0 be a homeomorphism. homeomorphism. Since Y has the universal extension property, property, there −1 is an extension of g of g given by h : Z → Y . composition given given by g ◦ h : Z → Y 0 Y . Then the composition gives the desired retraction. (b) Following the hint, we use the Tychonoff theorem to see that [0 , 1]J is normal, and then imbed Y in [0, [0, 1]J as given by theorem 34.2 (Imbedding theorem), as a normal space is completely regular. Let F : Y → RJ be the given imbedding. Let the indexed family of functions {f α }α∈J , f α : X → R satisfy that for every point x ∈ X and neighborhood U containing x, there is an index α such that f α is positive at x and vanishes outside U . U . Since Y is compact, a finite number of these U cover Y , Sincee Y , call them {U 1 , . . . , U2 }. Sinc J the imbeddings of these U i cover the imbedding of Y of Y in R , then Y is homeomorphic to a J J retract of R . Since we know that R and any retract of RJ have the universal extension property from Exercise 35.5(b), this implies Y has the universal extension property. Section 35, #7.
(a) The logarithmic spiral is homeomorphic to a ray coming out from the origin, and it intersects every circle centered at the origin precisely once (obvious by the formula, as et is a strictly increasing continous function onto R+ ). Thus our retraction should have every circle centered at the origin as an equivalence class, and the mapping r : R2 → C is given by r(x, y) = (et cos t, et sin t), where t = log x2 + y 2 for (x, (x, y) = (0, 0), and (0, (0 (0, 0). This part is continous, as the coordinates of r(x, y) are bounded below by of r r(0, , 0) = (0, (− x2 + y 2 , − x2 + y2 ) and above by ( x2 + y 2 , x2 + y 2 ). It maps all points to C , t t and given a point in C in the form (e ( e cos t, e sin t), we find log( x2 + y 2) = log(e log(et ) = t, as desired. (b) The knot K is homeomorphic to the real line R, wh whic ich h is a normal normal space space.. Th Then en by Excercises 35.5 and 35.6, since R has the universal extension property, it is a retract. Section 36, #1.
In a manifold X , every point x ∈ X has a neighborhood U that is homeomorphic to an open subset of Rm . As Rm is locally compact, so is each U . Exercise 33.7, every every locally U . From Exercise compact compact Hausdorff Hausdorff space is completely completely regular. regular. Our definition of manifold manifold already requires that it have a countable basis, so by the Urysohn metrization theorem, X is regular and has a countable basis imply X is metrizable.
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Section 36, #2.
Let U α be the neighborhoods of all the points x ∈ X , which can be imbedded in Rk for some Sincee X is compact, a finite number of these U α cover X , label the set k depending on x. Sinc U 1 , . . . , Un . X is Hausdorff and compact imply X is normal, so we may use Theorem 36.1 to show existence of a partition of unity dominated by {U i }. Then similarly to the proof of Theorem 36.2, we have the imbedding F : X → RN where N is finite because it is a finite product of finite numbers. Section 36, #3.
Every point in X has a neighborhood that is homeomorphic with an open subset of Rm . Morever, X compact and Hausdorff means we can use the previous exercise 36.2 to show that X can be imbedded in RN , which has a countable basis, thus X has a countable basis. So X is an m-manifold by the definition.
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