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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.

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# 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!