A short exact sequence is a kernel-cokernel pair

Assume the displayed sequence is short exact. Exactness at $A$ and [L4] make the arrow criterion [L2] read $1_{A}k=0$ for a kernel $k$ of $i$, so $\ker (i)=0$ and [L3] makes $i$ monic. Dually, exactness at $C$ gives $\mathrm{coker}(p)=0$, so $p$ is epic.

Conversely, assume $i$ is a kernel of $p$ and $p$ is a cokernel of $i$. Then [L7] makes $i$ monic and $p$ epic, so [L3] gives endpoint exactness. Also $pi=0$, and if $k$ is a kernel of $p$ while $q$ is a cokernel of $i$, then $k$ factors through $i$, hence $qk=0$ because $qi=0$. Thus [L2] gives exactness at $B$, so the sequence is short exact.

Exactness at $B$ gives $pi=0$ by [L2]. Since $i$ is monic by step 1.1, the factorization $i=i{1}_A$ is an epi-mono factorization of $i$, so every $u:U\to B$ with $pu=0$ factors through $i$. Thus $i$ is a kernel of $p$.

The same exactness at $B$, read through the second equality in [L5], identifies $\mathrm{coker}(i)$ with $\mathrm{coim}(p)$. Because step 1.1 makes $p$ epic, [L6] says that $p$ itself represents $\mathrm{coim}(p)$. Hence $p$ represents the quotient $\mathrm{coker}(i)$, which is exactly to say that $p$ is a cokernel of $i$.

Steps 1.1 and 2.1 show that short exactness forces $i$ to be the kernel of $p$.

Steps 2.2 and 1.2 complete the equivalence.