# Hatcher 0.11

Hatcher, Algebraic Topology, Chapter 0

11. Show that $f:X\to Y$ is a homotopy equivalence if there exist maps $g,h:Y\to X$ such that $fg\simeq 1$ and $hf\simeq 1$. More generally, show that $f$ is a homotopy equivalence if $fg$ and $hf$ are homotopy equivalences.

Proof. Suppose that $f:X\to Y$ is a map and that there exist maps $g,h:Y\to X$ for which $fg\simeq 1_Y$ and $hf\simeq 1_X$. The fact that $f$ is a homotopy equivalence (i.e., that it has a homotopy inverse) follows in numerous ways from these hypotheses. Here are a few.

• Because $fg\simeq 1_Y$, left composition with $h$ yields that $h(fg)=(hf)g\simeq 1_X\circ g=g$ and that this is homotopic to $h\circ 1_Y=h$. Hence, $g\simeq h$. In particular, then, $1_X\simeq hf \simeq gf$, which combined with the fact that $fg\simeq 1_Y$ yields the result.
• Because $hf\simeq 1_Y$, right composition with $g$ similarly yields that $g\simeq h$, whereby it follows that $1_Y\simeq fg \simeq fh$, which combined with $hf\simeq 1_X$ 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 $fg\simeq 1_Y$ implies that $hfg\simeq h$ \implies that $hfgf\simeq hf\simeq 1_X$ implies that $fhfgf\simeq f$. Tracing back through, this implies that $fhfg\simeq 1_Y\simeq fg$ and hence that $fh\simeq 1_Y$.

To show the more general result, suppose that $fg:Y\to Y$ and $hf:X\to X$ are homotopy equivalences with homotopy inverses $p:Y\to Y$ and $q:X\to X$, respectively. Then:

• By definition, $(fg)p=f(gp)\simeq 1_Y$ so the obvious candidate for a homotopy inverse here is $gp$. It suffices to show that $(gp)f\simeq 1_Y$. To that end, note that $p(fg)\simeq 1_Y$, whereby left composition by $g$ implies that $gpfg\simeq g$. In particular, then, $gpf=(gp)f\simeq 1_Y$, and so the result holds.
• Proceed as above: $q(hf)=(qh)f\simeq 1_X$ yields a candidate $qh$, and because $(hf)q\simeq 1_X$, right composition by $h$ shows that $hfqh\simeq h$. Hence, $fqh=f(qh)\simeq 1_X$ and the result again holds.

$\square$

## 4 thoughts on “Hatcher 0.11”

1. Bla

You need injectivity and/or surjectivity assumptions to use “cancellation laws” for composition of functions.

• Indeed, you are correct. Thanks for pointing out my absentmindedness!

2. Yilong Zhang

Thanks for your proof. Here is some typing error: Paragraph 4 the formula in last sentence “this implies that $fhgh \simeq 1_Y \simeq fg$” should be changed to $fhgh \simeq 1_Y \simeq fh$

The last sentence of that paragraph says: “…this implies that $fhfg\simeq 1_Y\simeq fg$ and hence that $fh\simeq 1_Y$.”