# Hatcher 0.5

Hatcher, Algebraic Topology, Chapter 0

5. Show that if a space $X$ deformation retracts to a point $x\in X$, then for each neighborhood $U$ of $x$ in $X$ there exists a neighborhood $V\subset U$ of $x$ such that the inclusion map $V\hookrightarrow U$ is nullhomotopic.1

Proof. Let $X$ be a topological space, let $x$ be an element of $X$ onto which $X$ deformation retracts, and let $U\subset X$ be an arbitrary neighborhood of $x$ in $X$. As per the statement of the problem, it suffices to find a neighborhood $V\subset U$ of $x$ for which the inclusion map $\iota:V\to U$ is nullhomotopic, meaning that there exists a homotopy connecting $\iota$ with a constant map.

Without loss of generality, suppose the family of maps $F:X\times I\to X$ for which $F(x,t)=f_t(x)$ represents the deformation retract of $X$ onto $x\in X$. In particular, then, $f_0=id_X$, $f_1(X)=x$, and $f_t(x)=x$ for all $t\in I$ and so it follows that $f_1$ is the constant map $X\to x$ and that $f_t$ fixes $x$ for all $t\in I$. Clearly, a deformation in this (strong) sense implies a deformation retract in the weak sense, so by previous results, there exists an inclusion $\iota:\{x\}\to X$ for which $f_1\iota\simeq 1_{\{x\}}$ and $\iota f_1\simeq 1_X$. Let $G:X\times I\to X$ be the homotopy connecting $\iota f_1$ to $1_X$ so that $G(x,0)=g_0(x)=1_X(x)=x$ and $G(x,1)=g_1(x)=(\iota\circ f_1)(x)$ for all $x\in X$.

Because $\{f_t\}_{t\in I}$ is a deformation retract of $X$ onto the point $x\in X$, the family $\{f_t\}$ can be thought of as a shrinking of $X$ onto the subset $\{x\}$, and because the family $\{f_t\}$ is continuous in both $t$ and $x$ (by definition of homotopy), it follows that there exists an element $T\in I$ for which $g_t^{-1}(U)\subsetneq U$ for $t\geq T$ and $g_t^{-1}(U)=U$ for $t. Let $V=g_T^{-1}(U)$. It suffices to show that the inclusion $V\hookrightarrow U$ is homotopic to a constant map.

To that end, write $t=kT$, $k\in[0,1]$, and define $H:U\times I\to V$ so that

$H(u,k)=h_k(u)=g_{kT}(x)$ for all $x\in X\cap U$2.

In particular, then, $h_0(u)=g_0(u)=u$ as $g_0=id_X$. Moreover, $h_1(u)=g_T(u)$ for all $u\in U$, and because $V=g_T^{-1}(U)$, $h_1(U)\subset V$. Finally, because $f_t$ is a (strong) deformation retract, because $G$ is defined in terms of $f_t$, and because $H$ is defined in terms of $G$, it’s easily confirmed that $h_t(V)\subset V$ for all $t$ and hence that $h_t$ is a weak deformation retract. Therefore, from the same problem cited above (see here), there exists an inclusion map $j:\{x\}\to V$ for which $h_1\circ j\simeq 1_{\{x\}}$. In particular, $h_1\circ j$ is null-homotopic.

At this point, all the ingredients are in place. The fact that the family $\{f_t\}$ homotopes $1_X\simeq \iota\circ f_1$ to $1_{\{x\}}$ implies that $\{f_t\}$ restricted to $V\subset U$ homotopes $1_V\simeq(\iota\circ f_1)|_V$ to $1_{\{x\}}$. Recall, however, that $f_1$ is the constant map $X\to\{x\}$, and so $(\iota\circ f_1)(V)=\iota(x)$. Thus, $\iota\simeq 1_V\simeq 1_{\{x\}}$. Combining this with the fact that $h_1\circ j\simeq 1_{\{x\}}$ yields that $h_1\circ j\simeq \iota$. Tracing through the chaos, the fact that $h_1=g_T$ is easily recovered, whereby $h_1\circ j\simeq \iota$ implies that the inclusion $V\hookrightarrow U$ is, indeed, homotopic to the constant map collapsing the entire domain to $x$. Hence, the result holds.   $\square$

1. Ideas for this proof were borrowed from solutions posted by Tarun Chitra to his personal webpage, http://www.tarunchitra.com. This particular problem was posted as part of the document: http://tarunchitra.com/papers/6510/hw5.pdf

2. The restriction “$x\in X\cap U$” is less-notationally-mind-numbing way of defining $h_t(u)=g_t|_U(x)$.

$H^{-1}(U)$ is a neighborhood of $\{x\} \times [0,1]$, so by compacity, it contains some cylinder $V \times [0,1]$ where $V$ is a neighborhood of $x$. For each $y \in V$, we have that $H(y,t) \in U$ for all $t$, so $H$ restricts to a homotopy : $V \times [0,1] \rightarrow U$ that verifies what we asked.