Five lemma

The Five lemma applies to a commutative diagram in an abelian category whose two rows are exact sequences in an abelian category:If $f_1$ is an epimorphism, $f_2$ and $f_4$ are isomorphisms, and $f_5$ is a monomorphism, then $f_3$ is an isomorphism. In particular, four surrounding isomorphisms force the middle one. Both the monomorphism and epimorphism properties of the second and fourth vertical morphisms enter the conclusion.^[ref]

For modules, injectivity follows by chasing an element $x\in A_3$ with $f_{3}x=0$. Commutativity and injectivity of $f_4$ give $a_{3}x=0$, so exactness gives $x=a_{2}u$. Then $b_2{f}_{2}u=0$, hence $f_{2}u=b_{1}v$. Surjectivity of $f_1$ gives $v=f_{1}w$, and injectivity of $f_2$ yields $u=a_{1}w$. Therefore $x=a_2{a}_{1}w=0$.

For surjectivity, take $y\in B_3$ and choose $z\in A_4$ with $f_{4}z=b_{3}y$. Since $f_5{a}_{4}z=b_4{b}_{3}y=0$, injectivity of $f_5$ and exactness give $z=a_{3}x$. Now $b_3(y-f_{3}x)=0$, so $y-f_{3}x=b_{2}u$ for some $u\in B_2$. Write $u=f_{2}v$. Commutativity gives $y=f_3(x+a_{2}v)$.

The weaker one-sided hypotheses $f_1,f_4$ epic and $f_2,f_5$ monic do not suffice. Over a nonzero field $k$, use the exact rows $0\to 0\to k\stackrel{1}{\to }k\to 0$ and $0\to k\stackrel{1}{\to }k\to 0\to 0$, with every vertical map zero. The first and fourth are epimorphisms, and the second and fifth are monomorphisms, but the middle zero map $k\to k$ is not an isomorphism.