Given a quasiisomorphism of DGAs $f:A\rightarrow B$ and a DG-module $M$ over $A$.
Is the canonical chain map  

\[M\rightarrow B\otimes\_A M \qquad m\mapsto 1\otimes m\]
an isomorphism on homology? 



Induction along a quasiisomorphism of DGAs
This is an elaboration on the L's comment. Let me use the cochain convention, so that the differential has degree +1.


Let $B=\mathbb{Z}/2$ (concentrated in degree 0) and let $A = \mathbb{Z}[e]/(e^2,de-2)$, where $|e|=-1$. Then the canonical projection $A\rightarrow B$ is a quasiisomorphism.


Let $M$ be the acyclic chain complex $\mathbb{Z}/2\rightarrow \mathbb{Z}/4\rightarrow \mathbb{Z}/2$ sitting in degrees $0,1,2$.


Since we want to equip this chain complex with an $A$-module structure, we have to figure out how $e$ could act. The Leibniz rule restricts the choices, for example if we chose multiplication with $e$ to be the zero map everywhere, the graded Leibniz rule would be violated at $\mathbb{Z}/4$.


Let $1\_i$ be the canonical generator of the $i$-th chain module of $M$. If we set
$e\cdot 1\_2 =2\cdot 1\_1$ ,$e\cdot 1\_1 =0$, we obtain an $A$-Module structure on $M$. 


Now if we tensor it with $\mathbb{Z}/2$ over $A$, we get the chain complex
\[\mathbb{Z}/2\stackrel{\cong}{\rightarrow }\mathbb{Z}/2\stackrel{0}{\rightarrow }\mathbb{Z}/2\]


which is clearly not acyclic and thus the canonical map cannot be an homology isomorphism.


PS. There is another module structure on $M$; given by $e\cdot 1\_2 =0$ ,$e\cdot 1\_1 =1\_0$ for which it works. 


