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Consider the following commutative diagram in an abelian category:

$$\require{AMScd}\begin{CD} @.0@.0@.0\\ @.@VVV@VVV@VVV\\ 0@>>>A@>>>B@>>>C@>>>0\\ @.@VVV@VVV@VVV\\ 0@>>>F@>>>G@>>>D@>>>0\\ @.@VVV@VVV@VVV\\ 0@>>>C@>>>D@>>>E@>>>0\\ @.@VVV@VVV@VVV\\ @.0@.0@.0 \end{CD}$$

where each row and column is a short exact sequence. Suppose that none of the morphisms in this diagram is zero morphism (aside from the obvious ones). If the middle row is split, can we conclude that the third row is also split? If not, what extra conditions do we need to impose for the third row to be split?

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  • $\begingroup$ It's at least true if the third column is also split. $\endgroup$ – Captain Lama Jun 8 '16 at 15:33
  • $\begingroup$ But the third column is the same as the third row. $\endgroup$ – Anonymous Jun 8 '16 at 15:55
  • $\begingroup$ Ah yes I did not notice this symmetry, sorry. $\endgroup$ – Captain Lama Jun 8 '16 at 15:56
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In an answer to a question you posted recently, a construction was given that satisfies all the conditions you want except that $A$, and therefore all the morphisms from it, are zero. You can easily adapt that to satisfy all your conditions by taking the direct sum with a diagram of the form $$\require{AMScd}\begin{CD} @.0@.0@.0\\ @.@VVV@VVV@VVV\\ 0@>>>W@=W@>>>0@>>>0\\ @.@|@|@VVV\\ 0@>>>W@=W@>>>0@>>>0\\ @.@VVV@VVV@VVV\\ 0@>>>0@>>>0@>>>0@>>>0\\ @.@VVV@VVV@VVV\\ @.0@.0@.0 \end{CD}$$

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