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 $a$ be in the kernel of $\gamma_3$ . Then $d(\gamma_3(a))=\gamma_4(da)=0$ , and $\gamma_4$ is injective, so $da=0$ . By exactness, $a=da'$ for some $a'\in A_4$ . Now, $d(\gamma_2a')=\gamma_3a=0$ , so $\gamma_2a'=db$ , and by the surjectivity of $\gamma_1$ , $b=\gamma_1a''$ . $da'''=\gamma_2^{-1}d\gamma_1(a'')=a'$ . Thus, $a=d^2a''=0$ . So, $\gamma_3$ is an injection.

Now, assume $b$ is not in the image of $\gamma_3$ . $db\neq 0$ , so $a'=\gamma_4^{-1}db\neq 0$ . $\gamma_5 da'=d^2b=0$ , and $\gamma_5$ is injective, so $da'=0$ , and there exists an $a''$ such that $da''=a'$ . Thus, $d(b-\gamma_3a'')=0$ . So there is an $\alpha$ such that $d\gamma_2 \alpha=b-\gamma_3a''$ . Thus, $\gamma_3(a''+d\alpha)=b$ . Thus, $\gamma_3$ is surjective.

This actually implies the result for all abelian categories, since by the Freyd embedding 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.