Do chain homotopic maps between dg-algebras induce “same” maps on dg-modules

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?

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.
• It might be added that the "correct" notion of homotopy is a dg-algebra map $h:A\to B[t,dt]$ with $t$ in degree zero, such that $h(a)(0)=f(a)$ and $h(a)(1)=g(a)$ for all $a\in A$. However I must admit I do not see the connection well. Seemingly it must hold that with such homotopy the answer becomes "yes" and that each "correct" homotopy gives rise to a chain homotopy between $f$ and $g$, while there is an obstruction to producing a "correct" homotopy from a chain homotopy. – მამუკა ჯიბლაძე Mar 20 '17 at 6:47
• @მამუკაჯიბლაძე I think that notion of homotopy is only good in characteristic zero (i.e. it is inherently commutative). The model I'd prefer for $A$ (quasi-)free on generators $G$ would be to take "$A\times I:= (F(G\oplus G\oplus G), d)$" and then look at maps from $A\times I$ to $B$. Then for more general $A$ I'd first replace $A$ with a resolution. – Gabriel C. Drummond-Cole Mar 20 '17 at 6:56