M8130 Algebraic topology, tutorial 01, 2017
23.2.2017
Remark. All sets are assumed to be topological spaces and all maps are assumed to be continuous unless stated otherwise. The symbol '=' will denote that two topological spaces are homeomorphic. The closed unit interval will be denoted by I or J.
Exercise 1. Prove that being homotopic is an equivalence relation (on the set of continuous maps between topological spaces).
Solution. Let /, g, k : X —> Y be such that / ~ g, g ~ k, i.e. there exist maps h,h' : X x I —?■ Y such that h(x, 0) = f(x), h(x, 1) = g(x), h'(x, 0) = g(x), h'(x, 1) = k(x).
• Reflexivity: the map Hi : X x I —>• Y defined by Hi(x,t) := f(x) for all t E I is a homotopy between / and itself.
• Symmetry: the map H2 : X x I —?■ Y defined by H2(x,ť) := h(x, 1 — t) for all t E I is a homotopy between g and /.
Transitivity: the map H-$ : IxI —y Y defined by Hs(x,t) : = is a homotopy between / and k.
h(x, 2t) for 0 < t < \ h'(x,2t) for|<i<l
□
Exercise 2. Let ~ be an equivalence relation on a topological space X. Prove that the map f : X/ ~—> Y is continuous ifffop-.X^Y is continuous, where p : X —> X/ ~ is the canonical quotient projection.
Solution. The direction "=^" follows from the facts that p is continuous (in fact, the quotient topology is the final topology with respect to p) and the composition of continuous functions is again continuous. For "<=", let U CY be open. Then
p-\rl{u)) = {foP)-\u)
is open by continuity of / o p, so f^iU) must also be open by the definition of quotient topology and we are done. □
Exercise 3. Show that Dn/Sn~1 = Sn using the map f : Dn —^ Sn given by
f(xi,.. .,xn) = (2a/1 - ||a:||a:,2 ||a;||2 - 1).
Solution. It's easy to see that / is continuous. Moreover, its restriction to the interior of Dn gives a bijection to S'n\{(0,..., 0,1)} (the inverse function is given by (y, z) H> -^===y) and
we have fiS*1-1) = {(0,... ,0,1)}, so we can define /' : Dn/Sn-1 -+ Sn by f([x])2= /(x). Then /' is a bijection, and by the previous exercise it is continuous. Finally, both Dn/Sn~1 and Sn are compact (Hausdorff) spaces (since both Dn and Sn are closed bounded subsets of W1 and Ka+1, respectively, and S'^1 C Dn is closed), so /' must be a homeomorphism (a general fact for continuous bijections between compact spaces). □
M8130 Algebraic topology, tutorial 01, 2017
23.2.2017
Exercise 4. Let f : X —>• Y and Mf = X x JU^xi Y. Moreover, let lx ■ X —> Mf be given by x ^ (x,0), iy : Y —> Mf be given by y ^ [y] and r : Mf —>• Y be given by r(y) = y, r(x,t) = f(x). Show that
i) Y is a deformation retract of Mf,
U) roLX = f,
Hi) tyo/~tj(.
Solution.
i) Geometrically, the deformation retraction is realized by pushing X along I towards Y.
ii) We have r o lx{x) = r(x, 0) = f{x) for all x G X.
in) The required homotopy h : X x J —>• Mf is given by h(x, s) = [(x, s)].
□
Exercise 5. Show that the pair (Mf,X) has the homotopic extension property (HEP), i.e. Lx is a cofibration.
Solution. Let g : / x J —>- {0} x JU/x {0} be any retraction such that g(0, s) = (0, s) and 0(1, s) = (1,0). Then the map r : Mf x J -> X x {0} x J U Mf x {0} defined by r(x, t, s) = (x, g{t, s)) and r(y, s) = (y, 0) is the required retraction. □
Exercise 6. The smash product between two based spaces is defined by
(C, Co) A (D, d0) := (C x D)/ (C x {d0} U {c0} x D). Show that Xj A A Y/B = (X x F)/(X xBUAxF).
Solution. Let pi : X x Y —> X/A x Y/B be given by p±(x,y) = ([x], [y]) and p2 : X/A x Y/B —> X/AAY/B be given by p2 ((N, [y])) = ([ [x] ], [ [y] ]). Then the composition p2°P\ is continuous and factors through (X xY)/(X x BLiAxY), which implies that the canonical bijection between (1x7)/(X xBUAx Y) and X/AAY/B is continuous (using exercise 2). Using the definition of quotient topology several times, it can be shown that this bijection is also open, hence a homeomorphism. □
Exercise 7. Let A = U {0}} C R. Show that (I, A) does not have the HEP, i.e. the inclusion A ^ I is not a cofibration.
Solution. If A x J U / x {0} was a retract of / x J, the retraction would have to preserve connected subsets. But Ax JU/x {0} is locally connected while fx J is not, a contradiction.
□