Four lemma in an abelian category

Write the top row as $W\stackrel{i}{\to }X\stackrel{j}{\to }Y\stackrel{p}{\to }Z$ and the bottom row as $W^{\mathrm{\prime }}\stackrel{i^{\mathrm{\prime }}}{\to }{X}^{\mathrm{\prime }}\stackrel{j^{\mathrm{\prime }}}{\to }{Y}^{\mathrm{\prime }}\stackrel{p^{\mathrm{\prime }}}{\to }{Z}^{\mathrm{\prime }}$. Assume that $\alpha$ and $\gamma$ are epic and that $\delta$ is monic. To prove that $\beta$ is epic, let $x^{\mathrm{\prime }}$ be a member of $X^{\mathrm{\prime }}$. Since $\gamma$ is epic, [L2] gives a member $y$ of $Y$ with $\gamma y\equiv j^{\mathrm{\prime }}{x}^{\mathrm{\prime }}$. Then $\delta py=p^{\mathrm{\prime }}\gamma y\equiv p^{\mathrm{\prime }}{j}^{\mathrm{\prime }}{x}^{\mathrm{\prime }}\equiv 0.$ Because $\delta$ is monic, [L1] gives $py\equiv 0$. Exactness of the top row at $Y$ now gives a member $x_0$ of $X$ with $j{x}_0\equiv y$ by [L3].

Assume instead that $\beta$ and $\delta$ are monic and that $\alpha$ is epic. To prove that $\gamma$ is monic, let $y$ and $y^{\mathrm{\prime }}$ be members of $Y$ with $\gamma y\equiv \gamma y^{\mathrm{\prime }}$. Then $\delta py=p^{\mathrm{\prime }}\gamma y\equiv p^{\mathrm{\prime }}\gamma y^{\mathrm{\prime }}=\delta p{y}^{\mathrm{\prime }},$ so [L1] gives $py\equiv p{y}^{\mathrm{\prime }}$. By [L4], replace $y$ and $y^{\mathrm{\prime }}$ by representatives on one common epic refinement of the witnesses for both equalities and define $t:=y-y^{\mathrm{\prime }}$. Then $\gamma t=pt=0$ and $y=y^{\mathrm{\prime }}+t$ on that domain. Exactness of the top row at $Y$ gives a member $x$ of $X$ with $jx\equiv t$ by [L3].

From step 1.1 we get $j^{\mathrm{\prime }}\beta x_0=\gamma j{x}_0\equiv \gamma y\equiv j^{\mathrm{\prime }}{x}^{\mathrm{\prime }}\mathrm{.}$ By [L4], replace $\beta x_0$ and $x^{\mathrm{\prime }}$ by representatives on a common epic domain and define $z^{\mathrm{\prime }}:=x^{\mathrm{\prime }}-\beta x_0$. Then $j^{\mathrm{\prime }}{z}^{\mathrm{\prime }}=0$ and $x^{\mathrm{\prime }}=\beta x_0+z^{\mathrm{\prime }}$ on that domain. Exactness of the bottom row at $X^{\mathrm{\prime }}$ gives a member $w^{\mathrm{\prime }}$ of $W^{\mathrm{\prime }}$ with $i^{\mathrm{\prime }}{w}^{\mathrm{\prime }}\equiv z^{\mathrm{\prime }}$ by [L3], and epicity of $\alpha$ gives a member $w$ of $W$ with $\alpha w\equiv w^{\mathrm{\prime }}$ by [L2]. Therefore $\beta (x_0+iw)\equiv \beta x_0+i^{\mathrm{\prime }}\alpha w\equiv \beta x_0+i^{\mathrm{\prime }}{w}^{\mathrm{\prime }}\equiv \beta x_0+z^{\mathrm{\prime }}\equiv x^{\mathrm{\prime }}\mathrm{.}$ So every member of $X^{\mathrm{\prime }}$ lifts along $\beta$, and [L2] makes $\beta$ epic.

From step 1.2 we get $j^{\mathrm{\prime }}\beta x=\gamma jx\equiv \gamma t\equiv 0.$ Exactness of the bottom row at $X^{\mathrm{\prime }}$ therefore gives a member $w^{\mathrm{\prime }}$ of $W^{\mathrm{\prime }}$ with $i^{\mathrm{\prime }}{w}^{\mathrm{\prime }}\equiv \beta x$ by [L3]. Because $\alpha$ is epic, [L2] gives a member $w$ of $W$ with $\alpha w\equiv w^{\mathrm{\prime }}$. Then $\beta (x-iw)\equiv \beta x-i^{\mathrm{\prime }}\alpha w\equiv \beta x-i^{\mathrm{\prime }}{w}^{\mathrm{\prime }}\equiv 0.$ Since $\beta$ is monic, [L1] yields $x\equiv iw$. Therefore $t\equiv jx\equiv jiw\equiv 0,$ because the top row is a complex. So $y\equiv y^{\mathrm{\prime }}$, and [L1] makes $\gamma$ monic.

Therefore the four lemma holds in both the epic and the monic form stated above.