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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 $$0 \to N \to \text{Con}(f) \to M[1] \to 0.$$

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  • $\begingroup$ Can't we just use the filtration on $N$, then continue with $N\oplus F^iM[1]$? $\endgroup$ Dec 19, 2013 at 9:33
  • $\begingroup$ Only if the filtration on $N$ is finite, right? $\endgroup$ Dec 19, 2013 at 18:50
  • $\begingroup$ Why? The filtration is still indexed by some ordinal... $\endgroup$ Dec 19, 2013 at 18:59
  • $\begingroup$ (To be clear, I mean take the filtration $0\subset F^1N\subset F^2N\subset\cdots \subset N\subset N\oplus F^1M[1]\subset \cdots$. I don't see why this is a problem even if the filtrations are infinite.) $\endgroup$ Dec 19, 2013 at 19:45
  • $\begingroup$ This seems problematic to me. In the definition the filtration is indexed by the first countable ordinal $\omega$, whereas your filtration is indexed by $\omega \cdot 2$, if I'm not mistaken. Is there some way to obtain an $\omega$-filtration from yours? $\endgroup$ Dec 19, 2013 at 21:14

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