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 x1 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.

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