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.   $\,\,\square$