Composing with an epimorphism preserves image in an abelian category Let
$$
A \overset{\pi}{\twoheadrightarrow}B \overset{f}{\to} C
$$
be maps in an abelian category $\mathcal{A}$, where $\pi$ is an epimorphism. I would like to show that $\operatorname{im}(f\circ \pi) \cong \operatorname{im} f$.
I can prove the result using the Freyd-Mitchell embedding theorem, but I would like a more elementary proof, using the axioms of abelian categories.
My proof using F-M Embedding
It is clearly true for $R$-modules. The embedding theorem tells us that there is a fully faithful, exact functor $F:\mathcal{A} \to R\mathrm{Mod}$ for some ring $R$. Then $F(f\circ \pi) = F(f)\circ F(\pi)$ and by exactness $F(\pi)$ is an epimorphism, so $\operatorname{im} F(f\circ \pi) = \operatorname{im} F(f)$.
Lemma: For any morphism $\alpha:X\to Y$ in $\mathcal{A}$, we have $F(\operatorname{im} \alpha) = \operatorname{im} F(\alpha)$.
Proof: By definition of $\operatorname{im}$ there are morphisms $X \overset{e}{\twoheadrightarrow} \operatorname{im} \alpha \overset{m}{\rightarrowtail}Y$ such that $m$ is mono and $m\circ e = \alpha$. Also, since abelian categories have all equalisers, $e$ is an epimorphism. This is taken by $F$ to a diagram
$$
F(X) \overset{F(e)}{\twoheadrightarrow} F(\operatorname{im}\alpha) \overset{F(m)}{\rightarrowtail} F(Y),
$$
where $F(m)$ is mono and $F(e)$ is epi since $F$ is exact, and $F(m)\circ F(e) = F(m\circ e) = F(\alpha)$. Since $F(m)$ is mono, up to isomorphism we may assume that $F(\operatorname{im} \alpha)$ is a submodule of $F(Y)$ and $F(m):F(\operatorname{im} \alpha) \hookrightarrow F(Y)$ is the inclusion.
Let $y \in \operatorname{im} F(\alpha)$. Then $y = F(\alpha)(x)$ for some $x \in F(X)$. But then $F(e)(x) = F(\alpha)(x) \in F(\operatorname{im}\alpha)$. So $\operatorname{im} F(\alpha) \subseteq F(\operatorname{im}\alpha)$.
Conversely let $y \in F(\operatorname{im}\alpha)$. Then since $F(e)$ is onto, $y = F(e)(x)$ for some $x \in F(X)$, which means that $y = F(\alpha)(x) \in \operatorname{im} F(\alpha)$. So $\operatorname{im} F(\alpha) = F(\operatorname{im}\alpha)$, as required.
$$\tag*{$\blacksquare$}$$
So finally we have that $F(\operatorname{im} (f\circ\pi)) = F(\operatorname{im} f)$, and it follows by full faithfulness of $F$ that $\operatorname{im}(f\circ\pi) = \operatorname{im} f$.
My Question
I find this proof to be pretty inelegant, especially as it uses the sledgehammer of F-M Embedding. Can anybody give me a simpler proof, especially one that just relies on the axioms of abelian categories? Thank you.
 A: The image of $f:B\to C$ is $\ker \newcommand\coker{\operatorname{coker}}\coker f$ (regarded as a subobject of $B$). So we want to show that $\ker \coker f = \ker \coker (f\pi)$ as subobjects of $B$. It suffices therefore to show that $\coker f \cong \coker (f\pi)$.
Now the cokernel of a morphism is the coequalizer of the morphism and $0$, so it suffices to show that in general, coequalizers are not affected by precomposition with epimorphisms.
Suppose we have a diagram $A\twoheadrightarrow B \rightrightarrows C$, where the epimorphism is $\pi$, and the parallel arrows are $f,g$. We want to show that $q:C\to Q$ is a coequalizer of $f$ and $g$ if and only if it is a coequalizer of $f\pi$ and $g\pi$.
This follows from the fact that $hf = hg$ if and only if $hf\pi = hg\pi$, since $\pi$ is an epimorphism. This tells us that we have a natural (in $D$) bijection
$$\{ h:C\to D \mid hf\pi = hg\pi \}\leftrightarrow  \{ h : C\to D \mid hf = hg\},$$
and $q:C\to Q$ is a coequalizer of $f$ and $g$ if and only if $q^*$ induces a bijection between maps $Q\to D$ and the set on the right, and $q$ is a coequalizer of $f\pi$ and $g\pi$ if and only if $q^*$ induces a bijection between maps $Q\to D$ and the set on the left, so this bijection establishes the equivalence.
