TITLE: Do chain homotopic maps between dg-algebras induce "same" maps on dg-modules
QUESTION [8 upvotes]: Let $A$ and $B$ be two dg-algebras over a field $k$. Let $f, g: A\to B$ be two maps between dg-algebras. We call $f$ and $g$ chain homotopic if there exists a degree $-1$ map $h: A\to B$ such that $f-g=d_Bh+hd_A$.
Now let $M$ be a right $A$ dg-module. For a given map $f: A\to B$, we could obtain a right $B$ dg-module by tensor product: $M\mapsto M\otimes_AB$. To avoid confusion we denote it by $M\otimes^f_AB$.

My question is: if $f, g: A\to B$ are chain homotopic, then is it true that $M\otimes^f_AB$ and $M\otimes^g_AB$ are quasi-isomorphic or further isomorphic up to homotopy? If not, is there any counter-example?

REPLY [4 votes]: If I understand your question correctly, the answer is no. 
Here is a counterexample. Let $A=k\langle 1,y\rangle$ with zero differential and trivial unital multiplication (put $y$ in degree $1$ for concreteness). Let $B=k[x]\langle 1, y\rangle$ with commutative multiplication and differential induced by $d(x) = y$. Then the inclusion $f:A\to B$ and the map $g:A\to B$ which sends $y$ to zero are both dg algebra maps, and are chain homotopic by $h(y)=x$.
Now let $M=k$ where $y$ acts trivially. Then $M\otimes_A^f B \cong k[x]$ with zero differential while $M\otimes_A^g B\cong B$, which is contractible.
Here are two related explanations for why this fails: 1) chain homotopy is not a good notion of homotopy for algebras and 2) projective or flat chain complexes are not necessarily projective or flat algebras.