Nine lemma in an abelian category

Assume the bottom row is short exact. Applying [L1] to the given diagram shows that the top row is exact at its first two nodes.

Write the horizontal maps as $a_1,a_2$, $b_1,b_2$, and $c_1,c_2$, and the vertical maps as $i_1,i_2,i_3$ and then $p_1,p_2,p_3$. It remains after step 1.1 to prove that $a_2$ is epic. Let $s$ be a member of $A_3$. Lift $i_{3}s$ along the epic map $b_2$ to a member $t$ of $B_2$. Exactness of the bottom row gives a member $u$ of $C_1$ with $c_{1}u\equiv p_{2}t$, and epicity of $p_1$ gives a member $y$ of $B_1$ with $p_{1}y\equiv u$. Then $p_2{b}_{1}y=c_1{p}_{1}y\equiv c_{1}u\equiv p_{2}t\mathrm{.}$ By [L4], pass to one common epic refinement of all the preceding equivalences and put $w:=t-b_{1}y$. Then $p_{2}w=0$, $t=b_{1}y+w$, and $b_{2}t=i_{3}s$ there. Exactness of the second column gives a member $x$ of $A_2$ with $i_{2}x\equiv w$. Consequently $i_3{a}_{2}x=b_2{i}_{2}x\equiv b_{2}w=b_{2}t\equiv i_{3}s\mathrm{.}$ Since $i_3$ is monic, $a_{2}x\equiv s$. Thus $a_2$ is epic by [L3], and the top row is short exact.

For the converse, pass to the opposite category. After drawing its vertical arrows downward, the original top row is the bottom row and the original bottom row is the top row. Thus the implication proved in steps 1.1 and 2.1, applied in the abelian category from [L2], carries short exactness of the original top row to short exactness of the original bottom row.

Hence, under the standing short-exactness of the middle row and all three columns, the top row is short exact if and only if the bottom row is short exact.