We consider the following commutative diagram with exact rows:
The five lemma states that if is epic, and and are monic, then is monic. By our chasing rules this means that for all , implies that . So let’s start with a member of and assume that .
First off we can tell that , and since is monic this means . Now the upper row is exact at , so there must exist a member with . We find , so by the exactness of the lower row at there must be a member with . Because is epic, there is a member with . Then , and so . Then , but exactness tells us that , which leads us to conclude that . Therefore is monic.
Dually, if and are epic while is monic we conclude that is epic. And then if is epic, and are isomorphisms, and is monic, then is an isomorphism.
It’s instructive to run through the above chase in both languages. First read as denoting an element of an abelian group, while denotes equality of elements. Then go back and reconstruct the proof in terms of morphisms between objects of general abelian categories. Thinking of members as behaving like actual elements allows us to create a proof that works for any abelian category we happen to be considering.