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One can find all over the internet that it is well-known (and obvious) that given a fiber bundle $F \to E \to B$, the equality $\chi(E) = \chi(F)\chi(B)$ holds ($\chi$ is the Euler characteristic). This is supposed to be true without any serious (= other than the Euler characteristics of $B$ and $F$ being defined) restrictions on $B$, $F$ and $E$. The thing is, I cannot find any reference for this statement.

What I know (and what I've found proof of) is that if $F \to E \to B$ is an orientable fibration, then this is indeed true. But how do I get rid of the orientability assumption?

I'd really appreciate any reference or a sketch of proof.

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    Have you (if you understand french math) checked Serre, Homologie singulère des espaces Fibrés? I'm pretty sure it's in there. EDIT: It is, but unfortunately in the oriented case too.2012-10-11

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I think, for finite simplicial complexes this is trival. Use the dual decomposition of the base space. The Euler characteristic of the preimage of each cell is that of the fibre. Now that of the total space can be obtained by taking the usual combinatorial formula for \chi (A_1 \cup ...\cup A_k).