First, assume that we are in modules over a ring (this is the most commonly used setting anyways).

The method of proof is what is usually called diagram-chasing.

Let be in the kernel of . Then , and is injective, so . By exactness, for some . Now, , so , and by the surjectivity of , . . Thus, . So, is an injection.

Now, assume is not in the image of . , so . , and is injective, so , and there exists an such that . Thus, . So there is an such that . Thus, . Thus, is surjective.

This actually implies the result for all abelian categories, since by the Freyd imbedding theorem, any abelian category is equivalent to a subcategory of modules over a ring. This trick is necessary since the trick above required us to have a notion of elements in the objects of our category, one which doesn't always make sense. The 5-lemma can be proved directly, but the proof is just less enlightening than the one above.