Image of a morphism in a category with zero object, equalizers and coequalizers Let’s assume that $\mathcal{C}$ is a category which has a zero object and equalizer and coequalizer of all the morphisms exist (so we have kernels and cokernels). Then, is it true that for an arbitrary morphism $f:A\to B$, $Im(f)=Ker(Coker(f))$? If not, is there any counterexample?
 A: I don't know why the other answer was deleted (it was almost correct), so here it is again: Consider the category $\mathbf{Rng}$ of rngs. It has a zero object, and it is complete and cocomplete. Images can be described as usual. The kernel of a rng homomorphism $\varphi : R \to S$ is the subrng $\{r \in R : \varphi(r)=0\} \to R$, and the cokernel of $\varphi$ is the quotient $S \to S/\langle \mathrm{im}(\varphi) \rangle$ by the ideal generated by the image of $\varphi$. Hence, $\ker(\mathrm{coker}(\varphi))=\langle \mathrm{im}(\varphi) \rangle$. Now there are many examples where the image is not an ideal in which case we have $\ker(\mathrm{coker}(\varphi)) \neq \mathrm{im}(\varphi)$. For example, when $R,S$ happen to be rings and $\varphi$ be a ring homomorphism, then $1 \in \mathrm{im}(\varphi)$ implies $\langle \mathrm{im}(\varphi) \rangle = S$, so $\mathrm{im}(\varphi)$ is an ideal iff $\varphi$ is surjective. So any non-surjective ring homomorphism gives an example.
A: A similar phenomenon, as in the nice example by Martin Brandenburg, can be observed in the category $\mathrm{AbTop_2}$ of Hausdorff topological abelian groups. It is an additive category (unlike $\mathbf{Rng}$), and given a morphism $G \stackrel{f}{\longrightarrow} H$ in $\mathrm{AbTop_2}$, the kernel and image of $f$ is defined as one would expect:
\begin{align*}
\ker (f) & = f^{-1} (0) \\
\mathrm{Im} (f) & = f(X)
\end{align*}
But the cokernel is defined as $H/\overline{f(X)}$ (the quotient needs to be Hausdorff after all). Thus, in $\mathrm{AbTop_2}$ we have $\overline{\mathrm{Im} (f)} = \ker \big ( \mathrm{coker} (f) \big)$ in general.
