Short five lemma in an abelian category

Assume that $f$ and $h$ are monic. Let $y:Y\to B$ and $y^{\mathrm{\prime }}:Y^{\mathrm{\prime }}\to B$ be members with $gy\equiv g{y}^{\mathrm{\prime }}$. By [L4], choose epimorphisms $u:T\to Y$ and $v:T\to Y^{\mathrm{\prime }}$ such that $gyu=g{y}^{\mathrm{\prime }}v$, and define the member $z:=yu-y^{\mathrm{\prime }}v:T\to B$. Then $gz=0$. Since $hpz=p^{\mathrm{\prime }}gz=0$ and $h$ is monic, [L1] gives $pz\equiv 0$. Exactness of the top row at $B$ now gives a member $x$ of $A$ with $ix\equiv z$ by [L3].

Since the bottom row is short exact, $i^{\mathrm{\prime }}$ is monic. From $i^{\mathrm{\prime }}fx=gix\equiv gz=0$ and the monicity of $i^{\mathrm{\prime }}$ and $f$, [L1] gives $x\equiv 0$, hence $z\equiv ix\equiv 0$. By [L4], after an epic refinement of $T$ the equality $z=yu-y^{\mathrm{\prime }}v=0$ is literal, so the resulting common epic representatives witness $y\equiv y^{\mathrm{\prime }}$. Thus [L1] makes $g$ monic.

If $f$ and $h$ are epic in the original diagram, then $f^{\mathrm{o}\mathrm{p}}$ and $h^{\mathrm{o}\mathrm{p}}$ are monic in the opposite abelian category. Applying steps 1.1 and 2.1 to the opposite morphism of short exact sequences makes $g^{\mathrm{o}\mathrm{p}}$ monic, so $g$ is epic.

If $f$ and $h$ are isomorphisms, they are in particular monic and epic. Steps 2.1 and 3.1 make $g$ both monic and epic, so [L5] makes $g$ an isomorphism.