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?



Multiplicativity of Euler characteristic for non-orientable fibrations
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.


