6
$\begingroup$

Let X be CW complex. I'm trying to prove (using Zorn's lemma) that there is maximal contractible subcomplex. Problem is that I'm not able to show that increasing union of contractible subcomplexes has to be contractible itself.

$\endgroup$

1 Answer 1

14
$\begingroup$

By various standard lemmas, a CW complex $X$ is contractible if and only if every map $u:S^{n-1}\to X$ (for any $n>0$) can be extended over $B^n$. In this context $u(S^{n-1})$ is compact and therefore (by another standard lemma) contained in some subcomplex with only finitely many cells. If $X$ is the union of some totally ordered family of subcomplexes $X_\alpha$, it follows that $u(S^{n-1})\subseteq X_\alpha$ for some $\alpha$. This is enough to prove what you want.

$\endgroup$
3
  • 1
    $\begingroup$ Calling Whitehead's theorem a standard lemma is a slight understating :) $\endgroup$ Oct 4, 2011 at 18:13
  • $\begingroup$ Saying "Whitehead's theorem" is almost as misleading (I believe Henry had proved at least two reasonably well-known results). $\endgroup$
    – Igor Rivin
    Oct 4, 2011 at 20:42
  • $\begingroup$ So for short: $\pi_n$ commutes with increasing unions in consideration and detect contractibility by Whitehead. $\endgroup$ Oct 4, 2011 at 21:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.