The Diagram Lemmas 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.

Because the outer maps are isomorphisms, they are monic and epic. Therefore [L1] shows that the middle map is monic and epic.

Applying [L2] to that middle map shows that it is an isomorphism.

Hence a morphism of short exact sequences with invertible outer maps has invertible middle map as well.

Form the pullback of $p^{\mathrm{\prime }}$ along $h$ shown above. Because $p^{\mathrm{\prime }}g=hp$, there is a unique comparison map $u:B\to P$ with $\alpha u=p$ and $\beta u=g$. Since $p^{\mathrm{\prime }}$ is the cokernel of $i^{\mathrm{\prime }}$ by [L1], it is epic, so [L3] makes the square also cocartesian, and [L2] identifies $\ker (\beta )$ with $\ker (h)$.

Assume $f$ and $h$ are monic. Then $\ker (h)=0$, so step 1.1 makes $\beta$ monic. If $ut=0$, then $pt=\alpha ut=0.$ Because $i=\ker (p)$ by [L1], there is $s$ with $t=is$. Now $0=\beta ut=gis=i^{\mathrm{\prime }}fs\mathrm{.}$ The map $i^{\mathrm{\prime }}$ is monic by [L1], and $f$ is monic by hypothesis, so $s=0$ and hence $t=0$. Thus $u$ is monic, and therefore $g=\beta u$ is monic.

Still with the pullback square of step 1.1, let $j:A^{\mathrm{\prime }}\to P$ be a kernel of $\alpha$; by [L2] this exists and agrees with the induced map from $\ker (p^{\mathrm{\prime }})=\mathrm{im}(i^{\mathrm{\prime }})$ to $P$. Assume now that $f$ and $h$ are epic. To show that $u$ is epic, let $t:P\to T$ satisfy $tu=0$. Then $tjf=tui=0.$ Since $f$ is epic, $tj=0$. The map $\alpha$ is epic as a pullback of the epic $p^{\mathrm{\prime }}$ by [L2], and hence is the cokernel of its kernel $j$ by [L5]. Therefore $t=r\alpha$ for a unique $r:C\to T$. But then $0=tu=r\alpha u=rp,$ and $p$ is epic by [L1], so $r=0$ and hence $t=0$. Therefore $u$ is epic.

To show that $\beta$ is epic under the same hypotheses, let $s:B^{\mathrm{\prime }}\to T$ satisfy $s\beta =0$. Since the square of step 1.1 is cocartesian by [L3], the compatible pair of maps $0:C\to T$ and $s:B^{\mathrm{\prime }}\to T$ induces a unique $v:C^{\mathrm{\prime }}\to T$ with $vh=0$ and $v{p}^{\mathrm{\prime }}=s$. Because $h$ is epic, $v=0$, so $s=0$. Thus $\beta$ is epic, and therefore $g=\beta u$ is epic.

If $f$ and $h$ are isomorphisms, steps 2.1 and 3.1 show that $g$ is both monic and epic. Therefore [L4] makes $g$ an isomorphism.

This gives the short five lemma again, now by a pullback-and-pushout argument and without any use of members.

Because the top row is short exact, [L1] says that $p$ is epic and $i=\ker (p)$. Form the pullback of $p$ along $k_h$:

Form the pushout of $q_f$ and $i^{\mathrm{\prime }}$:

Step 1.1 gives $j=\ker (\pi )$ and makes $\pi$ epic. By [L4], the sequence $0\to A\stackrel{j}{\to }P\stackrel{\pi }{\to }K\to 0$ is therefore short exact. Applying [L1] to this new short exact sequence shows that $\pi$ is also a cokernel of $j$.

The pullback relation gives $hp{\pi }^{\mathrm{\prime }}=h{k}_h\pi =0,$ so the snake-data square yields $p^{\mathrm{\prime }}g{\pi }^{\mathrm{\prime }}=0.$ Because $i^{\mathrm{\prime }}=\ker (p^{\mathrm{\prime }})$ by [L1], [L5] gives a unique map $a:P\to A^{\mathrm{\prime }}$ with $i^{\mathrm{\prime }}a=g{\pi }^{\mathrm{\prime }}\mathrm{.}$

Since ${\pi }^{\mathrm{\prime }}j=i$ and the left square commutes, we have $i^{\mathrm{\prime }}aj=g{\pi }^{\mathrm{\prime }}j=gi=i^{\mathrm{\prime }}f\mathrm{.}$ The map $i^{\mathrm{\prime }}$ is monic, so $aj=f$. Therefore $q_{f}aj=q_{f}f=0.$ Because $\pi$ is a cokernel of $j$ by step 2.1, [L5] yields a unique morphism $\delta :K\to Q$ with $\delta \pi =q_{f}a\mathrm{.}$

Composing with the pushout coprojection gives $\iota \delta \pi =\iota q_{f}a={\iota }^{\mathrm{\prime }}{i}^{\mathrm{\prime }}a={\iota }^{\mathrm{\prime }}g{\pi }^{\mathrm{\prime }},$ which is the required relation. If ${\delta }_1$ and ${\delta }_2$ both satisfy that relation, then $\iota {\delta }_1\pi =\iota {\delta }_2\pi \mathrm{.}$ Since $\pi$ is epic by step 1.1 and $\iota$ is monic by step 1.2, this forces ${\delta }_1={\delta }_2$.

Hence the connecting morphism exists and is unique.

By [L2], the induced kernel row is exact at $\ker (f)$ and $\ker (g)$, and the induced cokernel row is exact at $\mathrm{coker}(g)$ and $\mathrm{coker}(h)$. It remains to prove exactness at $\ker (h)$ and $\mathrm{coker}(f)$.

Let $k_h:K\to C$ be a kernel of $h$, $q_f:A^{\mathrm{\prime }}\to Q$ a cokernel of $f$, and $q_g:B^{\mathrm{\prime }}\to Q_g$ a cokernel of $g$. Use the pullback $P=B{\times }_{C}K$ from [L1], with projections ${\pi }^{\mathrm{\prime }}:P\to B$ and $\pi :P\to K$. That construction gives an epimorphism $\pi$, a map $a:P\to A^{\mathrm{\prime }}$ satisfying $i^{\mathrm{\prime }}a=g{\pi }^{\mathrm{\prime }}$, and $\delta :K\to Q$ satisfying $\delta \pi =q_{f}a$. Let $u:K_g\to K$ be the induced arrow from a kernel $k_g:K_g\to B$ of $g$, and let $c:Q\to Q_g$ be induced by $i^{\mathrm{\prime }}$. Thus $k_{h}u=p{k}_g$ and $c{q}_f=q_g{i}^{\mathrm{\prime }}$.

The pair $(k_g,u)$ induces $s:K_g\to P$ with ${\pi }^{\mathrm{\prime }}s=k_g$ and $\pi s=u$. Since $i^{\mathrm{\prime }}as=g{k}_g=0$ and $i^{\mathrm{\prime }}$ is monic, $as=0$. Consequently $\delta u=\delta \pi s=q_{f}as=0$.

Let $t:T\to K$ be a kernel of $\delta$. Pull back $\pi$ along $t$, obtaining $e_0:E_0\to T$ epic and $v:E_0\to P$ with $\pi v=t{e}_0$. Then $q_{f}av=\delta t{e}_0=0$. By [L3], $\mathrm{im}(f)=\ker (q_f)$, so $av$ factors through the image inclusion $m_f:I_f\to A^{\mathrm{\prime }}$. Factor $f=m_f{e}_f$ with $e_f:A\to I_f$ epic, and pull back $e_f$ along this factorization. The resulting $e_1:E\to E_0$ is epic and has a map $x:E\to A$ with $fx=av{e}_1$.

From $c{q}_f=q_g{i}^{\mathrm{\prime }}$ and $i^{\mathrm{\prime }}a=g{\pi }^{\mathrm{\prime }}$ one obtains $c\delta \pi =q_g{i}^{\mathrm{\prime }}a=q_{g}g{\pi }^{\mathrm{\prime }}=0$. Since $\pi$ is epic, $c\delta =0$. Now let $r:R\to Q$ be a kernel of $c$. Pull back the epimorphism $q_f:A^{\mathrm{\prime }}\to Q$ along $r$, obtaining an epimorphism $d_0:D_0\to R$ and $v:D_0\to A^{\mathrm{\prime }}$ with $q_{f}v=r{d}_0$. Then $q_g{i}^{\mathrm{\prime }}v=cr{d}_0=0$, so $i^{\mathrm{\prime }}v$ factors through $\mathrm{im}(g)=\ker (q_g)$. Pull back the epic image projection $B\to \mathrm{im}(g)$ along this factorization. This gives an epimorphism $d_1:D\to D_0$ and $y:D\to B$ with $gy=i^{\mathrm{\prime }}v{d}_1$.

Put $z={\pi }^{\mathrm{\prime }}v{e}_1-ix:E\to B$. Then $gz=i^{\mathrm{\prime }}av{e}_1-i^{\mathrm{\prime }}fx=0$, so $z=k_{g}w$ for a unique $w:E\to K_g$. Also $pz=p{\pi }^{\mathrm{\prime }}v{e}_1=k_{h}t{e}_0{e}_1=k_{h}uw$. Since $k_h$ is monic, $uw=t{e}_0{e}_1$. The composite $e_0{e}_1$ is epic; hence $\mathrm{coker}(u)t=0$. By [L5], $t$ factors through $\mathrm{im}(u)$. Together with $\delta u=0$ from step 2.1, this gives $\mathrm{im}(u)=\ker (\delta )$.

The bottom-row relation gives $hpy=p^{\mathrm{\prime }}gy=p^{\mathrm{\prime }}{i}^{\mathrm{\prime }}v{d}_1=0$, so $py=k_{h}x$ for a unique $x:D\to K$. The pair $(y,x)$ induces $s:D\to P$, and $i^{\mathrm{\prime }}as=gy=i^{\mathrm{\prime }}v{d}_1$ implies $as=v{d}_1$ by monicity of $i^{\mathrm{\prime }}$. Thus $\delta x=q_{f}as=q_{f}v{d}_1=r{d}_0{d}_1$. Because $d_0{d}_1$ is epic, $\mathrm{coker}(\delta )r=0$, so [L5] makes $r$ factor through $\mathrm{im}(\delta )$. Along with $c\delta =0$, this proves $\mathrm{im}(\delta )=\ker (c)$.

The two remaining nodes are exact by steps 3.1 and 3.2; the other nodes are exact by step 1.1. Therefore the snake sequence is exact.

Because the top row is exact and ends in $0$, the map $b$ is epic by [L1]. Because the bottom row is exact and begins at $0$, the map $k$ is monic by [L1]. Choose a kernel $k_{\gamma }:K\to Z$ of $\gamma$ and a cokernel $q_{\alpha }:U\to Q$ of $\alpha$. Form the pullback $P$ displayed above. By [L3], $\pi$ is epic. Since $l\beta {\pi }^{\mathrm{\prime }}=\gamma b{\pi }^{\mathrm{\prime }}=\gamma k_{\gamma }\pi =0,$ the kernel property of $k$ gives a unique map $r:P\to U$ such that $kr=\beta {\pi }^{\mathrm{\prime }}\mathrm{.}$

Let $j:J\to P$ be a kernel of $\pi$. By [L3], the induced map $J\to \ker (b)$ is an isomorphism. Exactness of the top row at $Y$ says that $\ker (b)$ is the image of $a$, so there is an epimorphism $e:X\to J$ with ${\pi }^{\mathrm{\prime }}je=a\mathrm{.}$ Then $krje=\beta {\pi }^{\mathrm{\prime }}je=\beta a=k\alpha ,$ and monicity of $k$ gives $rje=\alpha$. Therefore $q_{\alpha }rje=q_{\alpha }\alpha =0.$ Because $e$ is epic, $q_{\alpha }rj=0$. Since $\pi$ is epic, [L6] makes it the cokernel of its kernel $j$, so there is a unique morphism $\delta :K\to Q$ with $\delta \pi =q_{\alpha }r\mathrm{.}$

Applying [L4] to the given diagram gives exactness of $\ker (\alpha )\to \ker (\beta )\to \ker (\gamma )$ and of $\mathrm{coker}(\alpha )\to \mathrm{coker}(\beta )\to \mathrm{coker}(\gamma )\mathrm{.}$ If $a$ is monic, then the sequence $0\to X\stackrel{a}{\to }Y\stackrel{b}{\to }Z$ is exact, so the same theorem gives that $\ker (\alpha )\to \ker (\beta )$ is monic. Dually, if $l$ is epic, then $\mathrm{coker}(\beta )\to \mathrm{coker}(\gamma )$ is epic. Thus only exactness at $\ker (\gamma )$ and at $\mathrm{coker}(\alpha )$ remains.

Let $i:\ker (\beta )\to Y$ be a kernel of $\beta$, and let $s:\ker (\beta )\to K$ be the induced map with $k_{\gamma }s=bi$. Because $\beta i=0$, the pair $(i,s)$ factors through the pullback, giving $t:\ker (\beta )\to P$ with ${\pi }^{\mathrm{\prime }}t=i,\pi t=s\mathrm{.}$ Then $krt=\beta {\pi }^{\mathrm{\prime }}t=\beta i=0,$ so monicity of $k$ gives $rt=0$. Therefore $\delta s=\delta \pi t=q_{\alpha }rt=0,$ which proves that $\delta$ kills the image of $\ker (\beta )\to \ker (\gamma )$.

Let $u:T\to K$ be a kernel of $\delta$. Pull back $\pi$ along $u$, obtaining an epimorphism $e_0:E_0\to T$ and $v:E_0\to P$ with $\pi v=u{e}_0$. Then $q_{\alpha }rv=\delta u{e}_0=0$. By [L5], $\mathrm{im}(\alpha )=\ker (q_{\alpha })$, so $rv$ factors through the image inclusion of $\alpha$. Pull back the epic image projection $X\to \mathrm{im}(\alpha )$ along that factorization. This gives an epimorphism $e_1:E\to E_0$ and $x:E\to X$ with $\alpha x=rv{e}_1$.

Let $q_{\beta }:V\to Q_{\beta }$ be a cokernel of $\beta$, and let $c:Q\to Q_{\beta }$ be induced by $k$, so $c{q}_{\alpha }=q_{\beta }k$. Since $kr=\beta {\pi }^{\mathrm{\prime }}$, one has $c\delta \pi =q_{\beta }kr=q_{\beta }\beta {\pi }^{\mathrm{\prime }}=0$, and epicity of $\pi$ gives $c\delta =0$. Let $t:R\to Q$ be a kernel of $c$. Pull back the epimorphism $q_{\alpha }:U\to Q$ along $t$, giving an epimorphism $d_0:D_0\to R$ and $v:D_0\to U$ with $q_{\alpha }v=t{d}_0$. Then $q_{\beta }kv=ct{d}_0=0$, so $kv$ factors through $\mathrm{im}(\beta )=\ker (q_{\beta })$. Pull back the epic image projection $Y\to \mathrm{im}(\beta )$ to obtain an epimorphism $d_1:D\to D_0$ and $y:D\to Y$ with $\beta y=kv{d}_1$.

Put $z={\pi }^{\mathrm{\prime }}v{e}_1-ax:E\to Y$. Since $\beta z=krv{e}_1-k\alpha x=0$, it factors through a kernel $i:K_{\beta }\to Y$ of $\beta$ as $z=iw$. Let $s:K_{\beta }\to K$ be the induced arrow. Then $k_{\gamma }sw=biw=bz=b{\pi }^{\mathrm{\prime }}v{e}_1=k_{\gamma }u{e}_0{e}_1$. Monicity of $k_{\gamma }$ gives $sw=u{e}_0{e}_1$. Since $e_0{e}_1$ is epic, $\mathrm{coker}(s)u=0$. By [L6], $u$ factors through $\mathrm{im}(s)$; with $\delta s=0$ from step 3.2, this proves $\mathrm{im}(s)=\ker (\delta )$.

The square and bottom-row relation give $\gamma by=l\beta y=lkv{d}_1=0$. Thus $by$ factors through $k_{\gamma }:K\to Z$ as $by=k_{\gamma }x$ for a unique $x:D\to K$. The pair $(y,x)$ gives $w:D\to P$ with ${\pi }^{\mathrm{\prime }}w=y$ and $\pi w=x$. Since $krw=\beta y=kv{d}_1$ and $k$ is monic, $rw=v{d}_1$. Hence $\delta x=q_{\alpha }rw=q_{\alpha }v{d}_1=t{d}_0{d}_1$. The composite $d_0{d}_1$ is epic, so $\mathrm{coker}(\delta )t=0$. By [L6], $t$ factors through $\mathrm{im}(\delta )$; together with $c\delta =0$, this proves $\mathrm{im}(\delta )=\ker (c)$.

The two central nodes are exact by steps 4.1 and 4.2; the other nodes and the endpoint clauses follow from step 3.1. Thus the stated sequence is exact.

Regard each vertical arrow of the given ladder as an object of the arrow category ${\mathcal{A}}^{\to }$. Because kernels, cokernels, pullbacks, and pushouts in ${\mathcal{A}}^{\to }$ are computed componentwise, the entire ladder is again a Mac Lane snake diagram in ${\mathcal{A}}^{\to }$.

Applying [L2] in ${\mathcal{A}}^{\to }$ produces a connecting morphism between the arrow objects $\ker (\gamma )\to \mathrm{coker}(\alpha )\mathrm{.}$ Read componentwise in $\mathcal{A}$, that morphism is exactly the pair consisting of the two ordinary connecting morphisms together with the comparison square between them.

The defining square for the arrow-category connecting morphism commutes by construction, and uniqueness in [L2] forces that componentwise square to be the naturality square for the two ordinary connecting morphisms. Therefore the connecting morphism is natural.

Consider the morphism between the canonical split short exact sequences tikzcd 0 \arrow[r] & A \arrow[r, "j_A"] \arrow[d, "f"'] & A\oplus B \arrow[r, "\pi_B"] \arrow[d, "m"'] & B \arrow[r] \arrow[d, "g"'] & 0 \\ 0 \arrow[r] & B \arrow[r, "j_B"'] & B\oplus C \arrow[r, "\pi_C"'] & C \arrow[r] & 0 where, in biproduct matrix notation, $m=\left(\begin{array}{cc}f& -1_B\\ 0& g\end{array}\right),m(a,b)=(f(a)-b,g(b))\mathrm{.}$ The two squares commute, so [L1] applies.

The map $a\mapsto (a,f(a))$ identifies $\ker (gf)$ with $\ker (m)$: the equations $m(a,b)=0$ are exactly $b=f(a)$ and $gf(a)=0$. Dually, the map $B\oplus C⟶\mathrm{coker}(gf),(b,c)⟼[c+g(b)]$ identifies $\mathrm{coker}(m)$ with $\mathrm{coker}(gf)$. Under these identifications, the snake sequence of step 1.1 is exactly $0\to \ker (f)\to \ker (gf)\to \ker (g)\to \mathrm{coker}(f)\to \mathrm{coker}(gf)\to \mathrm{coker}(g)\to 0.$

The maps in step 2.1 are the canonical comparison maps of [L2], so the kernel-cokernel sequence of a composite is a special case of the snake lemma.

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.

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 exactness holds at $X$, $Y$, $X^{\mathrm{\prime }}$, and $Y^{\mathrm{\prime }}$, and that $\alpha$ and $\gamma$ are epic while $\delta$ is monic. Let $x^{\mathrm{\prime }}$ be a member of $X^{\mathrm{\prime }}$. By epicity of $\gamma$ and [L2], choose a member $y$ of $Y$ with $\gamma y\equiv j^{\mathrm{\prime }}{x}^{\mathrm{\prime }}$. Then $\delta py=p^{\mathrm{\prime }}\gamma y\equiv p^{\mathrm{\prime }}{j}^{\mathrm{\prime }}{x}^{\mathrm{\prime }}\equiv 0,$ so monicity of $\delta$ and [L1] give $py\equiv 0$. Exactness at $Y$ gives a member $x_0$ of $X$ with $j{x}_0\equiv y$ by [L3].

Assume instead that exactness holds at $X$, $Y$, $X^{\mathrm{\prime }}$, and $Y^{\mathrm{\prime }}$, that $\beta$ and $\delta$ are monic, and that $\alpha$ is epic. 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 at $Y$ gives a member $x$ of $X$ with $jx\equiv t$ by [L3].

Now $j^{\mathrm{\prime }}\beta x_0=\gamma j{x}_0\equiv \gamma y\equiv j^{\mathrm{\prime }}{x}^{\mathrm{\prime }}\mathrm{.}$ 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 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 [L2] makes $\beta$ epic.