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 $p: E \rightarrow B$ a fiber fundle with fiber $F$. I'd like to prove that if $U$ is a good-cover of $B$ then the Euler characteristic, that I denote with $\chi$ is $\chi(E)= \sum_{p,q}\sum_{\alpha_{0}, \cdots, \alpha_{p}} (-1)^{p+q}\dim H^{q}(p^{-1}(U_{\alpha_{0} \le \cdots le \alpha_{p}}))$. Then deduces from this fact that $\chi(E)=\chi(F)\chi(B)$.

share|cite|improve this question
Are you familiar with the Mayer-Vietoris spectral sequence for a covering? – Mariano Suárez-Alvarez Jan 17 '13 at 6:58
No a lot... but if you are accurate I can try to understand... – ArthurStuart Jan 17 '13 at 8:02
Please, try to make the title of your question more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. – Julian Kuelshammer Jan 17 '13 at 9:01
@ArthurStuart: I've merged your two accounts. You should be able to log-in with either set of credentials to the same account now. This way you can comment on and edit your own question. – Willie Wong Jan 17 '13 at 9:06
@ Mariano Suárez-Alvarez Could I find a simple proof of my claim somewhere? – ArthurStuart Jan 17 '13 at 9:20
up vote 1 down vote accepted

To amplify on Mariano's comment: If $\{U_i\}$ is an open cover of $X$, then there is a convergent Mayer-Vietoris spectral sequence $$ \bigoplus H^p(U_{i_1}\cap \cdots \cap U_{i_q}) \implies H^{p+q}(X).$$ Now use that the Euler characteristic of a spectral sequence, i.e. $\sum_{p,q} (-1)^{p+q}\dim E_{r}^{p,q}$, does not depend on $r$. This proves the first part.

If you don't know spectral sequences then I think you can also do this by the usual Mayer-Vietoris sequence for a cover with two open sets, and induction over the number of opens in your good cover. But I suspect this could be messier combinatorially.

For the second use that $p^{-1}(U_{i_1}\cap \cdots \cap U_{i_q}) \cong (U_{i_1}\cap \cdots \cap U_{i_q}) \times F$ and the Künneth theorem.

share|cite|improve this answer
What is $X$? I haven't understood how is made the Mayer-Vietoris spectral sequece – ArthurStuart Jan 17 '13 at 21:18
Where can I find a clean general formulation of the (co)homological MV spectral sequence? Weibel or McCleary doesn't have it. I'm guessing it goes like this: for any subsets $(U_i)_{i\in I}$ of a topological space $X$ such that their interiors cover $X$, there is a homological spectral sequence of $R$-modules with second page $E^2_{p,q}=???$ and converging to $H_{p+q}(X;R)$, and a cohomological spectral sequence of $R$-algebras with second page $E_2^{p,q}=???$ and converging to $H^{p+q}(X;R)$. Is this also true for other topological (co)homology theories? Does $I$ need to be countable? – Leon Jan 9 '14 at 1:24
Maybe $E_{p,q}^2=\bigoplus_{I'\subseteq I, |I'|=p}H_q(\bigcap_{i\in I'}U_i)$? Shouldn't there be unions of $U_i$ somewhere? – Leon Jan 9 '14 at 1:28

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.