TITLE: Multiplicativity of the Euler characteristic for fibrations
QUESTION [16 upvotes]: For a Serre fibration
$$
F\to E \to B ,
$$
with $F,E,B$ having the homotopy type of finite complexes, it is known that the Euler characteristic is multiplicative:
$$
\chi(E) = \chi(F)\chi(B) .
$$
However, if we more generally assume that $B$ and $F$ are finitely dominated spaces, then does multiplicativity hold as well? (Recall that a finitely dominated space is a retract of a homotopy finite one.)
If true I am looking for a reference. If false, please explain.
Added later: it's true if the base is homotopy finite since we can take the fiberwise double suspension to obtain a fibration with homotopy finite fibers having the same Euler characteristic.  So we only need to consider the case when the base is a finitely dominated and the fibers are homotopy finite.
Second Addition:  I can solve the problem in general if I can solve it in the following case:  Let  $\tilde B \to B$ be a finite, regular, $n$-sheeted covering space, where $B$ is finitely dominated. Then $\chi(\tilde B) = n\chi(B)$.
(Note that $\tilde B$ is again finitely dominated, since there is a finite covering $\tilde B\times S^1 \to B\times S^1$ and by a theorem of Mather, a space $X$ is finitely dominated if and only if $X\times S^1$ is homotopy finite. But since the base $B\times S^1$ is homotopy finite, we can put a finiteness structure on the total space as well.)
Third Addition: The case alluded to in my "Second Addition" holds, by the second answer I gave below.

REPLY [5 votes]: I have already written a lot, but Oscar Randal-Williams's comment below John Klein's second answer seems to really simplify the matter, so I decided to add an answer to that effect. I can remove any of the answers I have written if somehow they're "obfuscating the view" or anything.
This argument is just a simplification, that bypasses Pedersen-Taylor's result, and also the need for Swan's other theorem (suggested by Oscar) that $\tilde K_0(\mathbb Z[\pi])$ is finite for a finite group $\pi$ (I think the proof is harder, although I might be mistaken here). The argument is therefore due to Oscar and John in this respect, this is just my write-up of it.
Namely, as in all other answers reduce to the case of a finite covering space $p: E\to B$. Letting $F$ denote the fiber, and $Q$ the (finite) image of $\pi_1(B)\to Aut(F)$, let $f: B\to BQ$ denote the corresponding classifying map.
Define the following functor from perfect $\mathbb Z[\Omega B]$-modules to $\mathbb Q$-modules: $P\mapsto B_!( (P\otimes_\mathbb Z \mathbb Q) \otimes_\mathbb Q C_*(F;\mathbb Q))$.
The observation is that $C_*(F;\mathbb Q)$, as a $\mathbb Z[\Omega B]$-module, is of the form $f^* C_*(F;\mathbb Q)$, so this is $\mathrm{colim}_{BQ} f_!((P\otimes_\mathbb Z\mathbb Q)\otimes_\mathbb Q f^*C_*(F;\mathbb Q))\simeq \mathrm{colim}_{BQ} f_!(P\otimes_\mathbb Z\mathbb Q) \otimes C_*(F;\mathbb Q)$ (where $\mathrm{colim}_{BQ}$ is derived)
In other words, this functor factors as $Perf(\mathbb Z[\Omega B])\to Perf(\mathbb Z[Q])\to Perf(\mathbb Q[Q])\to Perf(\mathbb Q)$, but the most important part is the first part.
Indeed, $Q$ is finite, and so on $K$-theory, $K_0(\mathbb ZQ)\to K_0(\mathbb QQ)$ lands in the image of $K_0(\mathbb Q)$ (under the "induction" morphism) : this is (as far as I know) a theorem of Swan's (specifically, theorem 4.2 from his book "K-theory of finite groups and orders")
Now if $P= \mathbb Q[Q]^n$ is a free $\mathbb Q[Q]$-module, $\mathrm{colim}_{BQ}(P\otimes_\mathbb Q C_*(F;\mathbb Q)) = \mathrm{colim}_{BQ}(P\otimes_\mathbb Q C_*(F^{triv}; \mathbb Q)) = P_{hQ}\otimes_\mathbb Q C_*(F; \mathbb Q)$.
In particular, on $K$-theory, this composite has the same effect as $\mathrm{colim}_B (-)\otimes_\mathbb Q C_*(F;\mathbb Q)$, and so it sends the trivial coefficient system $\mathbb Z$ to $C_*(B;\mathbb Q)\otimes_\mathbb Q C_*(F;\mathbb Q)$, which of course is another name for $\chi(B)\chi(F) \in \mathbb Z\cong K_0(\mathbb Q)$.
But this map also sends this trivial coefficient system to $C_*(E;\mathbb Q)$, which is also another name for $\chi(E)\in \mathbb Z\cong K_0(\mathbb Q)$.
This proves the claim.