Free DG modules

Question

Let $A$ be a DG algebra and $f : F \to M$ a morphism of DG $A$-modules such that $F$ is free and the induced map $H^{\bullet}F \to H^{\bullet}M$ vanishes. Does it follow that $f$ is nullhomotopic?

My first thought: since $f$ maps $Z^{\bullet}F$ into $B^{\bullet}M$, we can lift $f[1] : F[1] \to M[1]$ along $d_M : M \to M[1]$ to obtain a map of graded modules $h : F[1] \to M$. The problem is that I don't see why $h \circ d_F = 0$, which is what we need for $dh = f$. Am I missing something?

Edit: Oops, if the differential on $A$ is nonzero then $d_M$ is generally not $A$-linear, so I guess we can't produce an $h$ in the way I described. But hopefully something like this is still true...

Answer 1

In retrospect my confusion came from the fact that $d_M$ isn't necessarily $A$-linear: just choose a basis $B \subset F$, lift $f(B) \subset B^{\bullet}M$ along $d_M$, and use this to define $h$.