# Hatcher 0.9

Hatcher, Algebraic Topology, Chapter 0

9. Show that a retract of a contractible space is contractible.

Proof. Let $X$ be a topological space which is contractible, $A\subset X$ a subspace, and suppose that $X$ retracts onto $A$ via the map $r$. To be more precise, suppose $x$ is a point and that $f:X\to \{x\}$ is a map with homotopy inverse $g:\{x\}\to X$ for which $fg\simeq 1_{\{x\}}$ and $gf\simeq 1_X$. Let $F:\{x\}\times I\to \{x\}$ (resp. $G:X\times I\to X$) denote the homotopy connecting $fg$ and $1_\{x\}$ (resp., connecting $gf$ and $1_X$). It suffices to find homotopies connecting $A$ to $\{x\}$ and vice versa.

Note that $r:X\to A$ being a retract implies immediately that the inclusion $\iota:A\to X$ satisfies $r\iota=\iota r= 1_A$. Solution of this problem, then, boils down to finding maps $h:A\to\{x\}$ and $k:\{x\}\to A$ so that the diagram in Figure 1 below commutes and for which $hk\simeq 1_A$, $kh\simeq 1_{\{x\}}$.

Figure 1

The obvious candidates here are $h=f\iota$ and $k=rg$, which by definition ensure that the diagram in Figure 1 commutes. Moreover,

$hk = (f\iota)(rg)=f(\iota r)g=fg\simeq 1_{\{x\}}$ and $kh=(rg)(f\iota)=r(gf)\iota\simeq r\iota=1_A$,

whereby it follows that $h$ and $k$ satisfy the desired properties with regards to the identity maps. Hence, the claim is proven.    $\square$

Note: The choices for $h$ and $k$ can be rectified with the “usual homotopy notation” by modifying the definitions of $F$ and $G$ given above simply by working relative to $A$ throughout. To be specific, let $F:\{x\}\times I\to\{x\}$ and $G:X\times I\to X$ be the homotopies mentioned above sending $fg\mapsto id_{\{x\}}$ and $gf\mapsto id_X$, respectively. In particular, $F(y,0)=(f\circ g)(y)$, $F(y,1)=x$, $G(y,0)=(g\circ f)(y)$, and $G(y,1)=y$ for all $y\in X$. Next, define homotopies $H:\{x\}\times I\to\{x\}$ and $K:A\times I\to A$ so that $H(y,t)=F(y,t)|_A$ and $K(y,t)=G(y,t)|_A$. It follows immediately, then, that $H(y,0)=F(y,0)=(f\circ g)(y)$, $H(y,1)=F(y,1)=x$, $K(y,0)=G(y,0)=(g\circ f)(y)$, and $K(y,1)=G(y,1)=y$ for all $y\in X\cap A=A$. Thus, $A$ has the same homotopy type as $\{x\}$ by way of the explicit homotopies $H$ and $K$.   $\square$