I have a commutative diagram of modules of the form $$\require{AMScd} \begin{CD} @. @VVV @VVV @VVV @VVV @. \\ ... @>>> A_n @>>> B_n @>>> C_n @>>> A_{n-1} @>>> ... \\ @. @VVV @VVV @VVV @VVV @. \\ ... @>>> D_n @>>> E_n @. F_n @>>> E_{n-1} @>>> ... \\ @. @VVV @VVV @VVV @VVV @. \\ ... @>>> G_n @>>> H_n @>>> I_n @>>> G_{n-1} @>>> ... \\ @. @VVV @VVV @VVV @VVV @. \\ ... @>>> A_{n-1} @>>> B_{n-1} @>>> C_{n-1} @>>> A_{n-2} @>>> ... \\ @. @VVV @VVV @VVV @VVV @. \\ \end{CD}$$ where each column is exact. Each row is exact as far as that's possible.
My question is whether there is some diagram-chasing magic that I can employ to obtain an arrow $E_n\to F_n$ that makes the DEF-row exact and the diagram commutative.
I think there should be such an arrow by analogy from short exact sequences. The BEH- and CFI-columns are exact triangles $H\to B \to E \to H[1] \to $ and $I\to C\to F\to I[1] \to$ respectively. If I had two short exact sequences $0\to H\to B \to E\to 0$ and $0\to I\to C\to F\to 0$ and morphisms left and center, then they would induce a morphism $E\to F$ that completes the diagram. My question is whether there are such completions in the derived category.
If it helps: In the specific example I'm thinking of I have additional information, namely $B_\ast = 0$ so that there are some isomorphism around one could use.
No, this does not work and I should have seen myself why. It kept looking at the wrong short exact sequences for comparison. Sure, any diagram of the form $$\require{AMScd} \begin{CD} 0 @>>> E @>>> H @>>> B @>>> 0 \\ @. @. @VVV @VVV @. \\ 0 @>>> F @>>> I @>>> C @>>> 0 \end{CD}$$ or of the form $$\require{AMScd} \begin{CD} 0 @>>> H @>>> B @>>> E @>>> 0 \\ @. @VVV @VVV @. @. \\ 0 @>>> I @>>> C @>>> F @>>> 0 \end{CD}$$ has a completion, but of course diagrams of the form $$\require{AMScd} \begin{CD} 0 @>>> B @>>> E @>>> H @>>> 0 \\ @. @VVV @. @VVV @. \\ 0 @>>> C @>>> F @>>> I @>>> 0 \end{CD}$$ do not in general, for example $$\require{AMScd} \begin{CD} 0 @>>> \mathbb{Z}/2 @>>> \mathbb{Z}/4 @>>> \mathbb{Z}/2 @>>> 0 \\ @. @VV=V @. @VV=V @. \\ 0 @>>> \mathbb{Z}/2 @>>> \mathbb{Z}/2\times\mathbb{Z}/2 @>>> \mathbb{Z}/2 @>>> 0 \end{CD}$$ has no completion.