# Isomorphic kernels imply pullback?

In Hilton/Stammbach's A Course in Homological Algebra, they are treating the Ext functor, and they give the following lemma:

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He implies (but doesn't say) that the same is not true if we replace equality with isomorphism. But I can't see why it wouldn't be. Do you agree, or have I missed something?

• How exactly does he imply that? – Eric Wofsey Dec 29 '15 at 3:46
• @EricWofsey He says that we are proving a "partial converse" to another theorem (which, as you might expect, says that an isomorphism in the first vertical map in a map of short exact sequences implies that the right square is a pullback)...also the fact that to my eyes it would have taken no extra effort to prove the more general version. (These don't sound like "implications" as I type them...I'm just confused why he did it this way, except to emphasize the case of equality in the context of the naturality of the Ext functor. – Eric Auld Dec 29 '15 at 6:48

You are right that the result is also true if you replace the equality by an isomorphism. The reason it is called "a partial converse" is that there is the extra assumption that $\nu'$ is an epimorphism; you don't need that to prove that a pullback square induces an isomorphism between the kernels (which is what Lemma 1.2 claims). This hypothesis is not (explicitly) present in Zimmerman, but that's a mistake : at the end of the proof he uses the short five lemma, but without the assumption that $\alpha$ is an epimorphism, the short sequences are not exact.