Multiplicativity of Euler characteristic for non-orientable fibrations Let $E\to B$ be a fibration with fiber F, and assume for simplicity that B is connected.  Suppose moreover that B and F have Euler characteristics (perhaps they are manifolds).  Then often, one can conclude that E has an Euler characteristic as well, and that
$$ \chi(E) = \chi(B)\cdot \chi(F). $$
The only proof of this that I have been able to find uses a spectral sequence argument, and requires that $\pi_1(B)$ act trivially on the homology of F, so that the homology in the spectral sequence can be taken with constant coefficients.  This condition is sometimes referred to as orientability of the fibration (with respect to the homology theory, normally rational homology).
Is the result known to be true any more generally than this?  Is there any other known proof?  Are there any examples where it is known to be false?
 A: Since this question seems to be attracting some renewed interest, I may as well point out that a few years after I asked it, Kate Ponto and I proved a generalization of this formula by purely homotopical/categorical methods, which applies to any fibration and yields a result about the Lefschetz number and even the Reidemeister trace.  The paper is here.
A: Assume for simplicity that $B,F$ are finite CW-complexes and let $p:E\to B$ be the bundle projection.
Suppose $B$ is obtained from a CW-complex $B'$ by attaching an $n$-cell. Suppose $\chi(B')\chi(F)=\chi (E')$ with $E'=p^{-1}(B')$. Then $H^*(E,E')\cong \tilde H(E/E')\cong H^*(D^n\times F,S^{n-1}\times F)$ [upd: some more details: $E/E'$ is the one point compactification of $p^{-1}(B\setminus B')$; now $B\setminus B'$ is an $n$-disk and so $p^{-1}(B\setminus B')\cong (D^n\setminus S^{n-1})\times F$, so the one point compactification of $p^{-1}(B\setminus B')$ is $(D^n\times F)/(S^{n-1}\times F)$. Now using excision and homotopy  we see that $\tilde H^*(E/E')\cong H^*(D^n\times F,S^{n-1}\times F)$.]
So $\chi(E,E')=(-1)^n\chi(F)$. So by induction on the number of cells we get $\chi(E)=\chi(B)\chi(F)$. No assumptions on the action of $\pi_1(B)$ are necessary.
A: Let $B$ be a finite simplicial complex. Let us consider its dual cell-decomposition.
The preimage of each cell has Euler characteristic equal to that of $F.$
Now use the usual combinatorial formula for the Euler characteristic of the union of finitly many sets, $\chi(A_1 \cup \dots \cup A_k)$, taking for $A_i$, the preimage of the dual cell of the $i$-th vertex of $B$. We obtain the required equality.
