Half nine lemma

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 second row is short exact, its left map is monic by [L1], so [L2] gives $i_{1}x\equiv i_{1}y$. The first column is also short exact, so its left map is monic; applying [L2] again yields $x\equiv y$. Hence the top-row map $A_1\to A_2$ is monic, so 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 second 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 short exact, its left map $c_1$ 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 short exact, $i_2$ is monic, so [L2] gives $a_{1}z\equiv t$. Thus every member of $A_2$ killed by $A_2\to A_3$ lifts from $A_1$, and the top row is exact at $A_2$ by [L3].

Therefore the top row is left exact whenever the bottom two rows and all three columns are short exact.