Nine lemma variants by which rows are assumed exact

If the bottom two rows are short exact, then the standing hypotheses of [L1] are met, so the top row is short exact.

If the top two rows are short exact, the same theorem [L1] applied after swapping the top and bottom rows shows that the bottom row is short exact.

Assume the top and bottom rows are short exact and that the middle row is a complex. To prove exactness at $B_1$, let $x$ be a member of $B_1$ with image $0$ in $B_2$. Applying the right map of the first column gives $c_1{p}_{1}x=p_2{b}_{1}x\equiv 0.$ Since the bottom row is short exact, its left map $c_1$ is monic by [L2], so [L3] gives $p_{1}x\equiv 0$. Exactness of the first column at $B_1$ gives a member $u$ of $A_1$ with $i_{1}u\equiv x$. Then $i_2{a}_{1}u=b_1{i}_{1}u\equiv b_{1}x\equiv 0.$ Because the second column is short exact, its left map $i_2$ is monic, so [L3] gives $a_{1}u\equiv 0$. The top row is short exact, hence its left map is monic by [L2]; another use of [L3] gives $u\equiv 0$, and therefore $x\equiv 0$. Thus the middle-row map $B_1\to B_2$ is monic.

Still under the same hypotheses, let $t$ be a member of $B_2$ with image $0$ in $B_3$. Because the bottom row is exact at $C_2$, there is a member $z$ of $C_1$ with $c_{1}z\equiv p_{2}t$ by [L3]. Since the first column is short exact, its right map is epic by [L2], so [L3] yields a member $x$ of $B_1$ with $p_{1}x\equiv z$. Then $p_2{b}_{1}x=c_1{p}_{1}x\equiv c_{1}z\equiv p_{2}t\mathrm{.}$ By [L3], pass to one common epic refinement of these equalities and the hypothesis $b_{2}t\equiv 0$, and define $w:=t-b_{1}x$. Then $p_{2}w=0$, $b_{2}w=0$, and $t=b_{1}x+w$ on that domain. Exactness of the second column at $B_2$ gives a member $u$ of $A_2$ with $i_{2}u\equiv w$ by [L3]. The middle row is a complex, so $i_3{a}_{2}u=b_2{i}_{2}u\equiv b_{2}w=0.$ Because the third column is short exact, $i_3$ is monic; [L3] gives $a_{2}u\equiv 0$. Exactness of the top row at $A_2$ therefore gives a member $v$ of $A_1$ with $a_{1}v\equiv u$ by [L3]. Hence $w\equiv i_{2}u\equiv i_2{a}_{1}v=b_1{i}_{1}v,$ so $t\equiv b_1(x+i_{1}v)\mathrm{.}$ This proves exactness of the middle row at $B_2$ by [L3].

Let $s$ be a member of $B_3$. Since the bottom row is short exact, its right map is epic by [L2], so [L3] gives a member $z$ of $C_2$ with $c_{2}z\equiv p_{3}s$. Since the second column is short exact, its right map is epic as well, choose a member $y$ of $B_2$ with $p_{2}y\equiv z$. Then $p_3{b}_{2}y=c_2{p}_{2}y\equiv c_{2}z\equiv p_{3}s\mathrm{.}$ By [L3], pass to one common epic refinement of these equalities and define $w:=s-b_{2}y$. Then $p_{3}w=0$ and $s=b_{2}y+w$ on that domain. Exactness of the third column at $B_3$ gives a member $u$ of $A_3$ with $i_{3}u\equiv w$ by [L3]. Because the top row is short exact, its right map is epic by [L2], so [L3] gives a member $v$ of $A_2$ with $a_{2}v\equiv u$. Therefore $w\equiv i_{3}u\equiv i_3{a}_{2}v=b_2{i}_{2}v,$ and hence $s\equiv b_2(y+i_{2}v)\mathrm{.}$ By [L3], the map $B_2\to B_3$ is epic.

Steps 1.3, 1.4, and 1.5 prove that the middle row is short exact.

These are exactly the three standard variants of the nine lemma distinguished by which rows are assumed exact.