Euler characteristic and universal cover Let $M$ be a compact manifold, let $\tilde{M}$ be its universal cover, and suppose that the Euler characteristic $\chi(\tilde{M})=0$.
My question is: does this imply that $\chi(M)=0$?
This is clear if $\pi_1(M)$ is finite, but I am interested in the case $|\pi_1(M)|=\infty$.
It might not feel right, but I can't think of any counterexample, either.
Thank you very much in advance!

EDIT: I was rightfully asked what I mean by Euler characteristic of the (non compact) manifold $\tilde{M}$. My answer right now is: the one you want!
What I am thinking of, is $\chi(\tilde{M})=\sum_i (-1)^i\dim H_i(\tilde{M},k)$, with $k=\mathbb{Q}$ or $\mathbb{R}$, and $H_i$ are either the usual or the compactly supported cohomology groups.
In my case, $\tilde{M}$ retracts to a compact Lie group.
 A: Here is sketch proof that $\chi(\tilde{M})=0$ implies that $\chi(M)=0$ for fundamental groups with certain finiteness properties.
The idea is to adapt the usual spectral sequence proof that the Euler characteristic is multiplicative on fibrations. Let $k$ be a field of characteristic zero, and let $\pi=\pi_1(M)$. The Cartan-Leray spectral sequence of the regular cover $\tilde M\to M$ has $E^2 = H_p(\pi;H_q(\tilde M;k))$ and converges to $H_{p+q}(M;k)$. But we can also go one page back and start at $E^1 =F_p\otimes_{k[\pi]} H_q(\tilde M;k)$, where $F_\bullet$ is a free resolution of $k$ by $k[\pi]$-modules. (Reference: K. Brown, Cohomology of groups, VII.5 and VII.7.) Now in a spectral sequence the Euler characteristic of each page is the Euler characteristic of the previous one, so it suffices to check that $\chi(E^1)=0$ when $\chi(\tilde M)=0$. 
If the trivial module $k$ admits a finite free resolution by $k[\pi]$-modules, we can choose each $F_p$ to be a finitely generated free $k[\pi]$ module, so $F_p$ is a direct sum $k[\pi]^{n(p)}$ for some $n(p)$. I think we then get $F_p\otimes_{k[\pi]} H_q(\tilde M;k)\cong k^{n(p)}\otimes_k H_q(\tilde M;k)$ as vector spaces, and the result follows in the usual way since the Euler characteristic of a tensor product of finite dimensional graded vector spaces is the product of their Euler characteristics.
A: Take a hyperbolic, compact, connected, oriented $3$-manifold $M$. Then  The universal cover $\tilde{M}$ is contractible and thus $\chi(\tilde{M})=\dim H_0(\tilde{M})=1$.
