MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 14 down vote accepted

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.

share|cite|improve this answer
Calling Whitehead's theorem a standard lemma is a slight understating :) – Mariano Suárez-Alvarez Oct 4 '11 at 18:13
Saying "Whitehead's theorem" is almost as misleading (I believe Henry had proved at least two reasonably well-known results). – Igor Rivin Oct 4 '11 at 20:42
So for short: $\pi_n$ commutes with increasing unions in consideration and detect contractibility by Whitehead. – Martin Brandenburg Oct 4 '11 at 21:54

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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