Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. 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 having only cells up to dimension $n$.

I want to prove that every $m$-dimensional vector bundle $E$ over $X$ decomposes as a sum $E\cong A\oplus B$ where $A$ is a $n$-dimensional (perhaps $n+1$?) vector bundle and $B$ is a trivial bundle $X\times \mathbb{R}^{m-n}$.

I remember vaguely that I've seen such a statement somewhere but I have no idea how to prove it. My first thought was to use the cellular approximation theorem an show that every map $X\to Gr(m,\infty)$ is homotopic to a map with image in $Gr(n,\infty)$ but I have doubts now that the cell structure on $Gr(n,\infty)$ can be choosen in such a way.

share|cite|improve this question
"My first thougt was to use..." -- yes, that's right, just do it! – Grigory M Feb 13 '12 at 20:38
I think your first thought is perfect. Milnor and Stasheff's book, Characteristic classes, theorem 6.4 provides a CW structrue for $Gr(n,\infty)$ with each of the $Gr(n,m)$ as finite subcomplexes. – Jason DeVito Feb 13 '12 at 20:39
Thank you for the responses. What should I do with the question now? – André Feb 13 '12 at 21:04
Post an answer, people will upvote it, then accept it! – user641 Feb 13 '12 at 23:13
I still don't get it. To use cellular approximation for the map $X\to Gr(m,\infty)$, $Gr(n,\infty)$ would have to be the $n$-skeleton of $Gr(m,\infty)$ but this isn't given by theorem 6.4, is it? – André Feb 14 '12 at 16:32

Your Answer


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

Browse other questions tagged or ask your own question.