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Let be $F \rightarrow E \rightarrow B$ be a fiber bundle where $B$ is path-connected and all three spaces are compact. Have you a simple proof that $\chi(E)=\chi(F) \chi(B)$? Where could I find that? I denote $\chi$ the Euler characteristic.

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What conditions on $F$ do you have? This doesn't hold in general, for example for $\mathbb Z \to \mathbb R \to S^1$ we'd have $1 = \chi(\mathbb Z) \cdot 0$. – Marek Jan 16 '13 at 19:27
up vote 5 down vote accepted

The proof that I know follows from the Serre spectral sequence. Let $k$ be a field, and suppose that $\pi_1(B)$ acts trivially on $H_*(F;k)$. (Clearly you also need to assume that $\chi(F)$ and $\chi(B)$ exist!). The $E_2$ term of the Serre spectral sequence becomes $$E_2^{s,t} \simeq H_s(B;k) \otimes H_t(F;k)$$

Now we need to define the Euler characteristic for the $r$-th page of the spectral sequence. This is $$\chi(E^r) = \sum_{s,t} (-1)^{s+t} \text{dim}E^r_{s,t}$$

So $\chi(E^2) = \chi(B)\chi(F)$.

Now one can show that $\chi(E^2) = \chi(E^3) = \cdots$, and since by assumption the spectral sequence collapses for large enough $r$ we have that $\chi(E^2) = \chi(E^\infty)$. Finally show that $\chi(E^\infty) = \chi(E)$. Putting this all together gives what you want.

For more details, see Spanier - Algebraic Topology, pp. 481-482

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Also… – Juan S Jan 17 '13 at 0:50

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