Sharp nine lemma

Write the horizontal maps of the three rows as $a_1,a_2$, $b_1,b_2$, and $c_1,c_2$, and the vertical maps of the three columns as $i_1,i_2,i_3$ from top to middle and $p_1,p_2,p_3$ from middle to bottom. Assume the three columns and the last two rows are exact at their first two nodes. Let $x$ and $y$ be members of $A_1$ with the same image in $A_2$. Commutativity gives the same image of $i_{1}x$ and $i_{1}y$ in $B_2$. Because the middle row is exact at its first node, $b_1$ is monic by [L1], so [L2] gives $i_{1}x\equiv i_{1}y$. Because the first column is exact at its first node, $i_1$ is monic, and another use of [L2] yields $x\equiv y$. Thus the top row is exact at $A_1$.

Let $t$ be a member of $A_2$ with image $0$ in $A_3$. Commutativity gives that $i_{2}t$ maps to $0$ in $B_3$. Exactness of the middle row at $B_2$ therefore yields a member $y$ of $B_1$ with $b_{1}y\equiv i_{2}t$ by [L3]. Applying the right map of the first column gives $c_1{p}_{1}y=p_2{b}_{1}y\equiv p_2{i}_{2}t\equiv 0.$ Because the bottom row is exact at its first node, $c_1$ is monic by [L1], so [L2] shows $p_{1}y\equiv 0$. Exactness of the first column at $B_1$ now gives a member $z$ of $A_1$ with $i_{1}z\equiv y$ by [L3]. Then $i_2{a}_{1}z=b_1{i}_{1}z\equiv b_{1}y\equiv i_{2}t\mathrm{.}$ Since the second column is exact at its first node, $i_2$ is monic, so [L2] gives $a_{1}z\equiv t$. Thus the top row is exact at $A_2$.

Assume in addition that the first column and the middle row are short exact. By steps 1.1 and 1.2, the top row is already exact at its first two nodes, so only epicity of $a_2:A_2\to A_3$ remains. Let $s$ be a member of $A_3$. Because the middle row is short exact, $b_2$ is epic by [L1], so [L2] gives a member $t$ of $B_2$ with $b_{2}t\equiv i_{3}s$. Then $c_2{p}_{2}t=p_3{b}_{2}t\equiv p_3{i}_{3}s\equiv 0,$ so exactness of the bottom row at $C_2$ gives a member $u$ of $C_1$ with $c_{1}u\equiv p_{2}t$ by [L3]. Since the first column is short exact, $p_1$ is epic by [L1], so [L2] gives a member $y$ of $B_1$ with $p_{1}y\equiv u$. Now $p_2{b}_{1}y=c_1{p}_{1}y\equiv c_{1}u\equiv p_{2}t\mathrm{.}$ By [L4], pass to one common epic refinement of all the preceding equivalences and define $w:=t-b_{1}y$. Then $p_{2}w=0$, $t=b_{1}y+w$, and $b_{2}t=i_{3}s$ on that domain. Exactness of the second column at $B_2$ gives a member $x$ of $A_2$ with $i_{2}x\equiv w$ by [L3]. Therefore $i_3{a}_{2}x=b_2{i}_{2}x\equiv b_{2}w=b_{2}t\equiv i_{3}s\mathrm{.}$ Because the third column is exact at its first node, $i_3$ is monic, so [L2] gives $a_{2}x\equiv s$. Hence $a_2$ is epic, and the top row is short exact.

Hence the sharp nine lemma is the left-exact half together with the precise extra hypotheses needed to upgrade it to a short exact row.