# Hatcher 0.12

Hatcher, Algebraic Topology, Chapter 0

12. Show that a homotopy equivalence $f:X\to Y$ induces a bijection between the set of path-components of $X$ and the set of path-components of $Y$, and that $f$ restricts to a homotopy equivalence from each path-component of $X$ to the corresponding path component of $Y$. Prove also the corresponding statements with components instead of path-components. Deduce that if the components of a space $X$ coincide with its path-components, then the same holds for any space $Y$ homotopy equivalent to.

Proof. Let $X,Y$ be topological spaces and let $f:X\to Y$ be a homotopy equivalence with homotopy inverse $g:Y\to X$. For notational convenience, let $P_X,P_Y$ denote the sets of path-components of $X,Y$ and define a map $\psi:P_X\to P_Y$ as follows: For every $x\in X$, let $[x]\in P_X$ denote the path-component of $X$ containing $x$1 and define $\psi$ to be the map2 for which $\psi:[x]\mapsto[f(x)]$. It suffices to show that this map is well-defined and that it indeed is a bijection as claimed.

To show well-definedness, suppose that $[x_0]=[x_1]$ in $P_X$, i.e. that $x_0$ and $x_1$ lie in the same path component of $X$. In particular, there is a continuous map $\gamma:[0,1]\to X$ for which $\gamma(0)=x_0$ and $\gamma(1)=x_1$. Consider the map $f\circ\gamma:[0,1]\to Y$. This map is continuous as it’s the composition of continuous maps, and it sends $f(\gamma(0))=f(x_0)$ to $f(\gamma(1))=f(x_1)$, meaning that $f(x_0)$ and $f(x_1)$ lie in the same path component in $Y$. Thus, $\psi([x_0])=[f(x_0)]=[f(x_1)]=\psi([x_1])$ and so the map $\psi$ is well-defined.

Finally, to show that $\psi$ is a bijection, it suffices to show that $\psi$ has a well-defined inverse $\eta:P_Y\to P_X$. Define $\eta:[y]\mapsto[g(y)]$ for all $[y]\in P_Y$. To show that these functions really are inverses, note that $\eta(\psi([x]))=\eta([f(x)])=[g(f(x))]$ and that, similarly, $\psi(\eta([y]))=[f(g(y))]$. By definition of homotopy inverse, there are homotopies $F:X\times I\to X$ and $G:Y\times I\to Y$ so that $F: gf\mapsto 1_X$ and $G:fg\mapsto 1_Y$. In particular, these homotopies defined “paths” connecting the maps $fg$ to $1_Y$ and $gf$ to $1_X$, whereby it follows that $g(f(x))$ and $f(g(y))$ are in the same path components as $x$ and $y$, respectively. Hence, $\eta\circ\psi:[x]\mapsto[x]$ and $\psi\circ\eta:[y]\mapsto[y]$ for all $x\in X, y\in Y$ and so it follows that $\psi, \eta$ are inverses.    $\square$

1. The notation $[x]$ for the path-component of $X$ containing $x\in X$ is suggestive of the presence of equivalence classes / quotient maps. Indeed, this stems from the fact that path-connectedness is an equivalence relation – a fact that’s well-known from point-set topology.

2. After floundering for a few hours trying to make my poorly-defined original map work, I stumbled upon this document online: http://www.stanford.edu/class/math215b/Sol1.pdf. Therefore, the idea for this particular definition belongs to him/her/them.