MATH 231: PSET 8

SITAN CHEN

1. Problem 1 (a) NOTE: I gured that itd help to make use of the fact that ev is continuous, but ultimately I couldnt gure it out and used the Internet to do some research. The below solution comes from http://neil-strickland.staff.shef.ac.uk/courses/homotopy/cgwh.pdf. Id be happy to try to reproduce this during oce hours to verify that (hopefully) I understand all the steps. EDIT: I saw Hiros email mentioning the lemma below the day after typing this up, oops. :P That Maps(X, Y ) is k -closed follows directly from the fact that the topology on Maps(X, Y ) is the k -ication of the compact-open topology. To prove that it is weakly Hausdor, we will use the following lemma: Lemma 1. A k -space X is weakly Hausdor i its diagonal inclusion X := {(x, x) | x X } is closed. This lemma would imply that the diagonal inclusion of Y is closed, and because any eval1 1 uation map evx is continuous, ev x evx is continuous so that the preimage of the diagonal

inclusion of Y , which is merely the diagonal inclusion of Maps(X, Y ), is closed. This implies that Maps(X, Y ) is indeed compactly generated. Proof. First assume the diagonal inclusion is closed. Then take f : K X and g : L X for compact Hausdor spaces f and g . Then dene space J = (f g )1 (X ), i.e. the subspace of K L consisting of pairs (k, l) where f (k ) = g (l). J is closed, and thus compact inside K L which is Hausdor. Because continuous maps take compact sets to compact sets, the projection map onto the L-coordinate takes J to a compact set and thus, by Hausdor-ness, a closed set. In particular, this closed set in L is the set of all l L for which there exists k K so that g (l) = f (k ). In other words, this is g 1 (f (K )). And because g 1 (f (K )) is closed, f (K ) is k -closed and thus closed.
Date : Collaborated with David Ding.
1

SITAN CHEN

In the other direction, assume X is weakly Hausdor. Therefore, X is weakly Hausdor so that X X is as well. As a result, it suces to show that for any map K X X given by (k K ) (f (k ), g (k )) from a compact Hausdorf K , the preimage of X is closed as well. This preimage is the set of k K such that f (k ) = g (k ). We wish to show that its complement is open, so pick any point in its complement, i.e. a point k such that f (k ) = g (k ). Take the set U of all k such that f (k ) = g (k ). This set contains k for sure, but in fact it is even an open neighborhood of k . After all, the set of all such k is the complement of the preimage of the (closed) single point subspace {g (k )} X under the map g , so it is open. Unfortunately, this U is too big, but because K is compact Hausdor, we can nd some open sub-neighborhood of a, call it V , whose closure is inside U . The image of its closure under g is closed, and f 1 (g (K V )) is thus open, and this gives us another neighborhood of a, call it W . We claim that V W is our desired open neighborhood of a contained inside the complement of (f g )1 (X ). We wish to show that for any x V W , f (x) = g (x). Because x W , f (x) g (V ). But because x V , g (x) g (V ) g (V ), meaning that f (x) = g (x), so we are done. (b) This follows from the homeomorphism between Maps(A, Maps(B, C )) and Maps(A B, C ) when A,B , and C are compactly generated, which is the case for A = [0, 1], B = X , and C = Y . If we have a homotopy between f0 and f1 , we have a continuous map F : [0, 1] X Y , and the homeomorphism proved in class tells us this corresponds to a continuous map f : [0, 1] Maps(X, Y ). Likewise, if we have a map f from [0, 1] to Maps(X, Y ) given by f (0) = f0 and f (1) = f1 , then our homeomorphism gives a continuous map F : [0, 1] X Y where F (x, t) = f (t)(x) so that we have a homotopy between f0 and f1 . 2. Problem 2 (a) It suces to show that (Mf , X ) has the the homotopy extension property, where we include X into Mf via X {0} Mf . By problem 1b on the fth problem set, it suces to show that Mf I retracts to ((X {0}) I ) (Mf {0}). We will exhibit this retraction geometrically: consider some the copy of the unit interval slightly behind X {1}, i.e. points ((x, 1), 1 + ). For each point ((x, s), t) inside Mf I , send this point along the ray starting at ((x, 1), 1) and going through this point to a point on (X {0}) I ) (Mf {0}). (see attached gure). In particular, points ((x, 1), 1) on X {1} modded out by identications to the same points in

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Y are sent by our map to ((x, 1), 0). It is clear that this map is continuous and the desired retraction. (b) From the previous problem we have a cobration from X to Mf , and Mf Y is a homotopy equivalence because the mapping cylinder of any map X Y deformation retracts onto Y . (c) (X, X k ) is a CW pair, so by problem set 5 exercises 3 and 4, it has HEP, so the inclusion is indeed a cobration. (d) Take the colimit to be the disjoint union of the point and Mf , quotiented out by the identication of points in the image of X {} and that of X Mf , i.e. the identication of the points in X {0} Mf , and this is just the mapping cone Cf . We know that any diagram

- Y

?
-

factors through

- Y

? - Cf
-

so that the colimit is indeed Cf .

3. Problem 3 (a) Say that the arrow Z Nf sends z to some (g (z ), h(z )), where g (z ) X and h(z ) Maps([0, 1], Y ). By commutativity of the square, if the homotopy Z [0, 1] Y is called F and homotopes some f0 = F (0, ) to some f1 = F (1, ), g (z ) = f0 (z ). We claim that the map G : Z [0, 1] X f Maps([0, 1], Y ) given by G(z, t) = (F (z, t), h(z )) is the dotted

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arrow. First check that this makes the diagram still commute: the upper left triangle consisting of Z , Z [0, 1], and Nf begins at z Z and in one direction sends to (z, 0) and then to (f0 (z ), h(z )) = (g (z ), h(z )), which is what z is sent to via Z Nf as well, so we have commutativity of that triangle. In the lower right triangle consisting of Z [0, 1], Nf , and Y , (z, t) is sent in one direction to (F (z, t), h(z )) and then to F (z, t), which is what (z, t) gets sent to via F : Z [0, 1] Y f as well, so we have commutativity there as well. It remains to show that this map is continuous. As a map from Z [0, 1] to the strict product X Maps([0, 1], Y ) this is continuous componentwise and thus continuous, and because we endowed Nf with the k -ication of the subspace topology inherited from this product, our map is continuous as a map to Nf as well. (b) Take the map Nf X given in the problem, which sends (x, ) to x, and call it . Also take the map : X Nf which sends x X to (x, f (x)), where f (x) denotes the constant path xed at point f (x). Obviously is the identity. We will prove that is homotopic to the identity. In particular, sends (x, ) to (x, f (x)). Our homotopy is F : Nf [0, 1] Nf dened by F ((x, ), s) = (x, s ) where s (t) = ((1 s)t). Then F ((x, ), 0) = (x, ) and F ((x, ), 1) = (x, f (x)) as desired, and this map is cerrtainly continuous. (c) This follows from a) and b) if we take N to be Nf . (d) Let the point that f takes to in Y be denoted by y Y . Then for any space X giving the diagram

Y 
6

Nf


 X our diagram factors through the following diagram:

MATH 231: PSET 8

Y 
6

Nf


 

X


X where X is dened to be the (x, ) Nf such that f (x) = x0 . This is the homotopy ber, which is presumably named because X can be thought of as the homotopies out of points in the ber of the base point of Y . 4. Problem 4 (a) Note that it suces to show that for spaces X , Y , and A X , MapsA (X, Y ) = Maps (X/A, Y ). From this, taking plugging in X Y , Z , and X Y for X , Y , and A, respectively, we get that Maps (X Y, Z ) = MapsX Y (X Y, Z ) = Maps(X Y, Z ) Maps(X Y, z0 ) = Maps(X, Maps(Y, Z )) Maps(X Y, z0 ). But the set of maps from X to Maps(Y, Z )) in this intersection are those which send basepoint x0 X to the constant map in Maps (Y, Z ) and which send all other x X to pointed maps in Maps (Y, Z ). Thus, the intersection is precisely Maps (X, Maps (Y, Z )). It remains to prove our claim that in general MapsA (X, Y ) = Maps (X/A, Y ). We will use the Yoneda lemma: it suces to prove that for any space Z , hom(Z, MapsA (X, Y )) = hom(Z, Maps (X/A, Y )). But hom(Z, Maps (X/A, Y )) = hom(Z X/A, Y ) = hom(XA , Maps(Z, Y )). But because X/A is the colimit of the diagram D A
- X

{} we then have the sequence of homeomorphisms hom(X/A, Maps(Z, Y )) = lim hom(D, Maps(Z, Y )) = lim hom(Z, Maps(D, Y )) = hom(Z, lim Maps(D, Y )).

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But the limit of Maps(D, Y ) is the limit of the diagram Maps(X, Y )

Maps({}, Y ) - Maps(A, Y ) But the limit of diagrams of this shape is the subspace of maps X Y whose restriction to A is the constant map, but this precisely the denition of MapsA (X, Y ), so we are done. (b) We have that [X, Z ] = [X, [S 1 , Z ]] = [X S 1 , Z ] = [, Z ]. The leftmost and rightmost equalities respectively come by denition of the based loop space of Z and the reduced suspension of Z . The isomorphism of sets in the middle comes from part a).

5. Problem 5 (a) We can use part b) of the previous problem. Recall that S n = S n+1 so that n S 0 = S n . From part b), we get an isomorphism of sets 0 (n Y ) = [S 0 , n Y ] = [n S 0 , Y ] = [S n , Y ] = n Y as desired. (b) Again, this follows merely from the fact that S n = S n+1 so that the left hand side is the set of homotopy classes of maps from S n to Y , which is precisely the denition of n Y . (c) If Y is connected, its zeroeth homotopy is zero. That Y is compactly generated tells us that a) holds and, by a similar argument, n1 ((Y )) = n (Y ), giving the desired result. As per the problem statement, we will merely observe that for n = 1, there is no immediately clear group structure on the zeroeth homotopy of Y . For n 2, however, we will demonstrate that our bijection preserves group structure so that we can recover not just the homotopy groups of either space from the other as sets, but as groups. We will show that group structure is respected from left to right in our bijection n1 (Y ) = n (Y ). Start with two classes [f ], [g ] n1 (Y ), i.e. homotopy classes of maps from S n1 to Maps (S 1 , Y ). Then [f ] [g ] is the homotopy class of the composite map S n1 S n1 S n1 Maps (S 1 , Y ) sending the top sphere in the wedge via f and the bottom sphere via g . This is sent by our bijection to the homotopy class of a map in n (Y ) (which is thought of as a map from (S n1 S n1 ) S 1 Y ) which sends x, y for x S n1 S n1 and y S 1 naturally to ((f g )(x))(y ). To show that this natural map is an isomorphism, we note the sequence of

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natural homeomorphisms Maps (X Y, Maps (Z, W )) = Maps ((X Y ) Z, W ) = Maps ((X Z ) (Y Z ), W ). Taking X and Y to be S n1 , Z to be S 1 , and W to be Y , we are done.

6. Problem 6 (a) For each point in the image of the path, pick an open neighborhood whose preimage in E is a disjoint union of open sets homeomorphic to the neighborhood. By compactness of the unit interval, nitely many of these cover the image of the path. Label these as U1 , ..., Un as follows: split the unit interval up into closed intervals Ik = [tk1 , tk ] for k = 1, ...m where t0 = 0 and tn = 1 and where (Ik ) Uk for all k . To construct the unique lift of onto E , we proceed inductively along these tk in the unit interval. For I1 , pick among the connected components of
1 1 p1 (U1 ) the one, call it V1 , which contains x so that |I1 has to be given by p 1 , where p1 is

the homeomorphism U1 V1 given by (p|V1 )1 . Now proceed by induction: say that we have constructed (the unique) |k Ii for some k . Then for Ik+1 , among the connected components
1

of

p1 (Uk+1 )

1 the Vk+1 which contains (tk ), and then |Ik+1 has to be given by p k+1 , where

1 : [0, 1] E in the p k+1 is the homeomorphism Uk+1 Vk+1 . In this way, weve gotten a lift

only way possible, giving us both existence and uniqueness. (b) A covering map of a path-connected space B is a ber bundle and thus a Serre bration: the only thing that needs to be shown is that for each b B , p1 (Ub ) consists of the same number, call it n, of connected components each homeomorphic to Ub , because then the ber of the ber bundle is just the discrete space of n points and the bundle map is given by homeomorphing each of the n components of any p1 (Ub ) via p to a copy of Ub . To show the uniformity property of covering maps of path-connected spaces, say for the sake of contradiction that there are two points x, y B with neighborhoods Ux and Uy which are pulled back to m and n connected components {Vk } and {Wk } homeomorphic to Ux and Uy respectively, where m > n. Then because B is path connected, consider a path from x to y . By part a), for each k = 1, ..., m, we have a unique path lifting which starts at the preimage of x inside Vk . Each of these lifted paths ends up in Wk for some k . By the pigeonhole principle there will be two lifted paths i and j for some i and j which end up in the same Wk , and say they are paths starting in Vi and Vj . But this means we have a path

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between connected components Vi and Vj , namely : [0, 1] E given by (t) = i (2t) for t [0, 1/2] and j (2t 1) for t [1/2, 1], a contradiction because by composition of paths we now have a path from every point in Vi to every point in Vj ! (c) Fix n, the degree of the disk out of which we have a map Dn E in our diagram, as well as a map Dn I B . As stated, we already know that a homotopy lifting exists; we merely need to show its uniqueness. And to do this, we only need to show that if we have two homotopy 1 and F 2 , then their restrictions to {x} I for all x Dn agree. But if we x any liftings F such x, we merely want to show that the following diagram has a unique dotted arrow:

{x} Dn

- E -

F
{x} I Dn I
?

p
? - B

But {x} I = I so that uniqueness of the dotted arrow follows from our result of part a), and were done. 7. Problem 7 (a) By the long exact sequence of brations, we have an exact subsequence n (F ) n (E ) n (B ) n1 (F ). Recall in our proof of 6b) that the ber F is a discrete space of points so that n (F ) = 0 for n 1. Therefore, for n 2, we get the isomorphism n (E ) = n (B ). (b) R and S 1 are both path-connected and locally contractible, so n (S 1 ) = n (R) = 0. (c) Because p1 and p2 are covering spaces, E1 B1 and E2 B2 are surjective so that the product of these two maps E1 E2 B1 B2 is likewise surjective. It remains to show that for every point (b1 , b2 ) B1 B2 , there is a neighborhood whose preimage is the union of open connected components each homeomorphic to the neighborhood. But there is a neighborhood in B1 of b1 , call it U1 , such that (p1 )1 (U1 ) is the disjoint union of connected components {Vn } each homeomorphic to U1 , and there is neighborhood in B2 of b2 , call it U2 , such that (p2 )1 (U2 ) is the disjoint union of connected components {Wn } each homeomorphic to U2 . Thus, we can simply take the desired neighborhood of (b1 , b2 ) to be U1 U2 so that its preimage in p1 p2 is

MATH 231: PSET 8

{Vi Wj } for all pairs i, j , and the components of each of these are homeomorphic to U1 and U2 respectively. (d) Because R S 1 is a covering map, the product of k such maps Rk T k is a covering map. We thus get an isomorphism n (Rk ) n (T k ). However, n (Rk ) = 0 for n 2. (e) Because E is contractible and thus path-connected, p : E X into a single path-connected component of X , call it X . However, p is a covering space and thus surjective, meaning X must be this single path-connected component. So by part a), we have n (E ) = n (X ) for n 2, and because E is contractible, n (E ) = 0 so that for n 2, n (X ) = 0. By problem 4 of the previous problem set, which tells us that the homotopy group of a product is isomorphic to the product of the homotopy groups of the components, n (T k ) = n (S 1 ) n (S 1 ).