Sorry for my ignorance in advance, this should be a very naive question and I would be happy for a reference. Let $G$ be an arbitrary group (not necessary finite) acting on two (connected) manifolds $M$ and $N$. The action on $M$ is on the left and the action on $N$ is on the right. Suppose that the action
$M\times G\rightarrow M$ is free. Is it true that $$H_{\ast}(M\times_G N, k)= H_{\ast}(M/G \times N,k )$$ where $k$ is a field or $\mathbb{Z}$ and the topology on $M/G$ is defined by the quotient map $M\rightarrow M/G$.

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    $\begingroup$ What is true geometrically is that because of the freeness of the action on $M$ the obvious map $M\times_GN\to M/G$ is a fiber bundle with fibers isomorphic to $N$. In general the homology of a bundle does not depend just on base and fiber, and Grodal's answer gives an example. A simpler one is the Klein bottle ($G$ of order two, $M=N=S^1$). $\endgroup$ – Tom Goodwillie May 16 '15 at 17:18

No, this is in general quite far from being true (and also somewhat ill-posed, since there is not a natural candidate for what "=" means since there is no map).

For example, take $M = S^1$ and $N=S^2$, both with the antipodal $C_2$-action (which you can view as either right or left action).

Then $M/C_2 \simeq S^1$ so $H_*(M/C_2 \times N;k) \cong H_*(S^1 \times S^2;k)$ as graded groups.

But $C_2$ acts as $-1$ on $H_2(N)$ and trivially on $H_*(M)$ so $H_*(M \times_{C_2} N;{\mathbb Q}) \cong H_*(M \times N;{\mathbb Q})_{C_2} \cong H_*(M;{\mathbb Q}) \cong H_*(S^1;{\mathbb Q})$ as graded abelian groups, where $C_2$ on the outside denotes coinvariants.

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