0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Your CD tex doesn't seem to be working $\endgroup$ – Alex G. Dec 24 '15 at 16:23
  • $\begingroup$ Strange. It works fine for me. (although the preview had some difficulties) $\endgroup$ – Johannes Hahn Dec 24 '15 at 16:27
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.