= Solution
In an <abelian category>, for $f:A\to B$ set $\operatorname{coim}f=\operatorname{coker}(\ker f)$ and $\operatorname{im}f=\ker(\operatorname{coker}f)$. The canonical map from coimage to image is an <isomorphism>. Thus the <image factorization in an abelian category> is
$$
\boxed{A\xrightarrow{e}\operatorname{im}f\xrightarrow{m}B,\qquad f=me,\quad e\text{ epic},\quad m\text{ monic},}
$$
unique up to a unique compatible <isomorphism>.
The <short five lemma> states that in a commuting diagram of <short exact sequences> $0\to L\to M\to R\to0$ and $0\to L'\to M'\to R'\to0$ in an <abelian category>, if the vertical maps $L\to L'$ and $R\to R'$ are <isomorphisms>, then so is $M\to M'$.
For the <Five lemma>, consider commuting exact rows
$$
A_1\xrightarrow{d_1}A_2\xrightarrow{d_2}A_3\xrightarrow{d_3}A_4\xrightarrow{d_4}A_5,
\qquad
B_1\xrightarrow{d'_1}B_2\xrightarrow{d'_2}B_3\xrightarrow{d'_3}B_4\xrightarrow{d'_4}B_5,
$$
with vertical maps $v_i:A_i\to B_i$. We prove the usual stronger version: $v_1$ epic, $v_2$ and $v_4$ invertible, and $v_5$ monic imply that $v_3$ is invertible. In particular the conclusion holds if all four outer vertical maps are <isomorphisms>.
Put $L=\operatorname{im}d_2$ and $R=\operatorname{im}d_3$, and define $L',R'$ similarly. Exactness gives <short exact sequences>
$$
0\longrightarrow L\longrightarrow A_3\longrightarrow R\longrightarrow0,
\qquad
0\longrightarrow L'\longrightarrow B_3\longrightarrow R'\longrightarrow0.
$$
The <functoriality of abelian image factorization> supplies their vertical outer maps. We show that both are <isomorphisms>.
First, $L\cong\operatorname{coker}d_1$ and $L'\cong\operatorname{coker}d'_1$: exactness at $A_2$ identifies $\operatorname{im}d_1$ with $\ker d_2$, and the image-coimage <isomorphism> for $d_2$ gives the claimed cokernel. Write $q:A_2\to L$ and $q':B_2\to L'$ for these <categorical cokernels>. To construct an inverse to the induced map $\ell:L\to L'$, observe
$$
(qv_2^{-1}d'_1)v_1=qv_2^{-1}v_2d_1=0.
$$
Cancellation of the <epimorphism> $v_1$ gives $qv_2^{-1}d'_1=0$, so $qv_2^{-1}$ factors uniquely as $\bar\ell q'$ with $\bar\ell:L'\to L$. From $\ell q=q'v_2$, composing with the epic $q,q'$ gives $\bar\ell\ell=1_L$ and $\ell\bar\ell=1_{L'}$.
Second, $R\cong\ker d_4$ and $R'\cong\ker d'_4$ by exactness at the fourth objects. Write their inclusions as $j:R\to A_4$ and $j':R'\to B_4$. The equality $d'_4v_4=v_5d_4$ gives the induced map $\rho:R\to R'$. Since
$$
v_5d_4v_4^{-1}j'=d'_4j'=0
$$
and $v_5$ is a <monomorphism>, $d_4v_4^{-1}j'=0$. Kernel universality gives $\bar\rho:R'\to R$ with $j\bar\rho=v_4^{-1}j'$. Cancelling the monic $j,j'$ proves $\bar\rho\rho=1_R$ and $\rho\bar\rho=1_{R'}$.
Apply the <short five lemma> to the two <short exact sequences>: their outer maps $\ell,\rho$ are invertible, so \b[$v_3$ is an <isomorphism>]. This <five lemma via image factorization> argument uses only universal properties and therefore works in any <abelian category>, without treating its objects as literal sets of elements.
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