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.



4 thoughts on “Hatcher 0.11

  1. 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 $fhgh \simeq 1_Y \simeq fh$

    • Thanks for your comment!

      Ignoring the fact that was pointed out above (namely, that all of my ‘cancellations’ were nonsense as written), I think your comment may have a typo and that my entry may have tons of typos. ^_^

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

      And, looking back through: I feel like my entire entry is malformed. I’m not really in Hatcher mode (and haven’t been for some time) so it’s unlikely I’ll even attempt to rectify one/any/all of the changes needed here for (at least) quite some time. As it stands, though, it’s very possible that there are typos abound, so I would suggest you proceed with (at worst) caution (and at best complete disregard for this entry’s existence) before attempting to reconcile anything here whatsoever.

      Thanks for reading, and for taking the time to comment.

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