Hatcher 0.3

Hatcher, Algebraic Topology, Chapter 0

3.  (a) Show that the composition of homotopy equivalences X\to Y and Y\to Z is a homotopy equivalence X\to Z. Deduce that homotopy equivalence is an equivalence relation.  (b) Show that the relation of homotopy among maps X\to Y is an equivalence relation.  (c) Show that a map homotopic to a homotopy equivalence is a homotopy equivalence.

Proof. Recall that a map f:X\to Y is called a homotopy equivalence if there exists a map g:Y\to X such that fg\simeq 1 and gf\simeq 1. In this case, g will be referred to as the homotopy inverse of f. Also, one says that two maps f_0,f_1:X\to Y are homotopic if there exists a homotopy f_t connecting f_0 to f_1. Here, one writes f_0\simeq f_1.

(a) Note that the main claim here is that homotopy equivalence is a transitive relation, the proof of which will verify that homotopy equivalence is an equivalence relation due to the fact that 1_X:X\to X is a homotopy equivalence (where 1_X is its own homotopy inverse), and that X is homotopy equivalent to Y if and only if Y is homotopy equivalent to X.

To that end, suppose that X,Y,Z are topological spaces and that f_1:X\to Y, f_2:Y\to Z are homotopy equivalences with homotopy inverses g_1,g_2, respectively. By definition, then, f_1g_1\simeq 1_X, g_1f_1\simeq 1_Y and f_2g_2\simeq 1_Y, g_2f_2\simeq 1_Z. It suffices to show that the map f_2f_1:X\to Z is a homotopy equivalence, which is immediate due to the fact that it’s continuous (the composition of continuous maps is continuous) and that g_1g_2:Z\to X is a map for which

(f_2f_1)\circ(g_1g_2)=f_2\circ(f_1g_1)\circ g_2\simeq f_2\circ 1_Y\circ g_2 = f_2g_2\simeq 1_Z

and

(g_1g_2)\circ(f_2f_1)=g_1\circ(g_2f_2)\circ f_1 \simeq  g_1\circ 1_Z\circ f_1 = g_1f_1\simeq 1_X.

Hence, the result.

(b) To show that the relationship of homotopy among maps X\to Y is an equivalence relation, note again that the reflexive and symmetric properties are free. Indeed, if f,g:X\to Y are maps which are homotopic by way of a homotopy \varphi_t, then f\simeq f by way of the identity homotopy F(x,t)=x and g\simeq f by way of the homotopy \varphi_t^{-1}. So, again, it suffices to prove that the relation is transitive.

To that end, suppose that F:X\times I\to Y is a homotopy connecting f to g and that G:X\times I\to Y is a homotopy connecting g to h, where f,g,h:X\to Y. Define, then, a map H:X\times I\to Y which takes on the value F(x,t) for 0\leq t\leq 1/2 and the value G(x,t) for 1/2< t \leq 1. Continuity of H is immediate due to the fact that H is made continuous at t=1/2 (the only point of concern), whereby it follows that f\simeq h.

(c) Finally, suppose that f:X\to Y is a homotopy equivalence with homotopy inverse g:Y\to X and suppose that h:X\to Y is homotopic to $f$, i.e. that there exists a homotopy F:X\times I\to Y connecting h to f. Said a different way, h\simeq f says that there exists a family \{f_t:X\to Y\}_{t\in I} connecting h to f, a fact which immediately implies that the family \{g\circ f_t\}_{t\in I} connects g\circ h to g\circ f\simeq 1_X. Thus, because 1_Y\simeq f\circ g by the homotopy inverse property, and because f\circ g\simeq h\circ g because h\simeq f\iff f\simeq h by (b), it follows that h:X\to Y is also a homotopy equivalence.   \,\,\square

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