# Prove an isomorphism via abstract nonsense

Suppose we are working in an abelian category and we have a commutative diagram with exact rows


with $f_1$, $f_3$ and $f_4$ isomorphisms. Can we then conclude (by diagram chasing maybe?) that $f_2$ is also an isomorphism?

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This follows trivially from five lemma, –  xyzzyz Nov 14 '13 at 11:44
Doesn't this follow from the five lemma? –  Alex Youcis Nov 14 '13 at 11:44
To add a little bit more to the last two comments - add two extra zeros on the left, and an isomorphism between the two you already drew. Then you can use the five lemma. –  Matthew Pressland Nov 14 '13 at 11:45
Interesting to know that \uparrow and \downarrow can be used for \left and \right! –  Lord_Farin Nov 14 '13 at 12:46

Thanks for pointing it out, it's indeed an immediate application of the 5 lemma. As Matt said, just add a pair of zeros on the left to get the commutative diagram


Now it's clear that the five lemma implies that $f_2$ is an isomorphism.

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