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A colleague and I have recently found ourselves using the following result:

$\require{AMScd} \begin{CD} @. @. 0 @. @. \\ @. @. @VVV @. @.\\ @. @. A @. @. \\ @. @. @VV\phi_1 V @. @. \\ 0 @>>> D @>>\psi_1> B @>>\psi_2> E @>>> 0\\ @. @. @VV\phi_2 V @. @. \\ @. @. C @. @. \\ @. @. @VVV @. @. \\ @. @. 0 @. @. \end{CD}$

If the two sequences here are exact, then $\ker(\psi_2 \circ \phi_1) \cong \ker(\phi_2\circ\psi_1)$, and similarly for cokernel.

The proof is straightforward diagram chasing, but we'd like to just cite a source for it if possible. He says he's seen a version of this referred to as the "cross lemma", but a search on this name doesn't yield much besides https://link.springer.com/content/pdf/10.1007/BF00969298.pdf and https://core.ac.uk/download/pdf/82599979.pdf, both of which are working in confusingly more general contexts.

Is there a basic source on algebra where we could find this written down? Does it go by a different name?

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  • $\begingroup$ Is it valid in abstract Abelian categories too? $\endgroup$ – Lehs Aug 4 '17 at 18:26
  • $\begingroup$ @Lehs Yes, because the diagram chasing in question works in any abelian category. Alternatively, one can use the argument I give below. $\endgroup$ – user144221 Aug 6 '17 at 11:53
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What you are looking for is a corollary of [a certain variation of] the $9$-lemma (also known as the $3\times 3$-lemma); the name "cross lemma" might not be so widely used in the literature.

You may consider the factorizations of $\psi_2\circ \phi_1$ and $\phi_2\circ\psi_1$ through their respective images, and complete the arrows $\phi_1$, $\psi_1$ to a square by taking a pullback, and $\phi_2$, $\psi_2$ by taking a pushout. Then you obtain a diagram with exact rows and columns

$\require{AMScd} \begin{CD} @. 0 @. 0 @. 0 @. \\ @. @VVV @VVV @VVV @.\\ 0 @>>> \bullet @>>> A @>>> \bullet @>>> 0 \\ @. @VVV @VV\phi_1 V @VVV @. \\ 0 @>>> D @>>\psi_1> B @>>\psi_2> E @>>> 0\\ @. @VVV @VV\phi_2 V @VVV @. \\ 0 @>>> \bullet @>>> C @>>> \bullet @>>> 0 \\ @. @VVV @VVV @VVV @. \\ @. 0 @. 0 @. 0 @. \end{CD}$

I believe there are some modern references for this statement, but it appears precisely in this form as Proposition 16.5 of Mitchell's "Theory of Categories" (1965).

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