Hatcher, Algebraic Topology, Chapter 0
11. Show that is a homotopy equivalence if there exist maps such that and . More generally, show that is a homotopy equivalence if and are homotopy equivalences.
Proof. Suppose that is a map and that there exist maps for which and . The fact that is a homotopy equivalence (i.e., that it has a homotopy inverse) follows in numerous ways from these hypotheses. Here are a few.
- Because , left composition with yields that and that this is homotopic to . Hence, . In particular, then, , which combined with the fact that yields the result.
- Because , right composition with similarly yields that , whereby it follows that , which combined with yields the result again.
Note that this is hardly the only two ways to deduce the same claim and, indeed, crazy combinations of compositions yield the same thing. For example implies that \implies that implies that . Tracing back through, this implies that and hence that .
To show the more general result, suppose that and are homotopy equivalences with homotopy inverses and , respectively. Then:
- By definition, so the obvious candidate for a homotopy inverse here is . It suffices to show that . To that end, note that , whereby left composition by implies that . In particular, then, , and so the result holds.
- Proceed as above: yields a candidate , and because , right composition by shows that . Hence, and the result again holds.