Let $A$ be a DG algebra. Recall that an $A$-module $M$ is free if it is isomorphic to a sum of shifts of $A$, or semi-free if it has an exhaustive filtration $0 = F_0 \subset F_1 \subset \cdots \subset M$ such that each $F_i/F_{i+1}$ is free.
Here's my question: is the cone of a morphism of semi-free modules semi-free? In Drinfeld's "DG quotients of DG categories" he claims this without proof.
I know how to prove that the cone of two semi-free modules is a direct summand of a semi-free module: an $A$-module $M$ is a summand of a semi-free module if and only if for any surjective quasi-isomorphism $N \to P$, the induced map $$\text{Hom}^{\bullet}(M,N) \to \text{Hom}^{\bullet}(M,P)$$ is a surjective quasi-isomorphism. But the latter property is clearly stable under extensions.
Edit: Maybe the cone construction is not as standard as I thought, so I'll give it here. If $f : M \to N$ is a morphism of $A$-modules, then $\text{Con}(f) = M[1] \oplus N$ as a graded module with the differential $$ \begin{bmatrix} -d_M & 0 \\ f[1] & d_N \end{bmatrix}.$$ The cone is a homotopy cokernel in the sense that for any $A$-module $P$, a morphism $\text{Con}(f) \to P$ is the same as a morphism $N \to P$ together with a nullhomotopy of the composition $M \to N \to P$. Note that there is a natural short exact sequence $$0 \to N \to \text{Con}(f) \to M[1] \to 0.$$