Is it true that $\ker(\operatorname{coker}\varphi)$ satisfies the universal property of images in a pre-abelian category?

Question

Let $\varphi:A\to B$ be a morphism in a pre-abelian category (that is, an additive category where every morphism has a kernel and a cokernel) $\sf{C}$. If $\sf{C}$ is indeed abelian, then I know that $i=\ker(\operatorname{coker}\varphi):K\to B$ satisfies the following universal property: $\varphi$ factors through $i$ and if $L\to B$ is another monomorphism through which $\varphi$ also factors, then it exists a unique morphism $K\to L$ such that the diagram

commutes. I want to know if $\ker(\operatorname{coker}\varphi)$ satisfies this universal property even if $\sf{C}$ is only pre-abelian.

I don't think this is the case since this universal property "gives" a morphism $\operatorname{im} \varphi\to\operatorname{coim}\varphi$ (given that we define $\operatorname{im}\varphi$ as the source of $\ker(\operatorname{coker}\varphi)$ and $\operatorname{coim}\varphi$ as the target of $\operatorname{coker}(\ker\varphi)$) and not the other way around. This is ok if $\operatorname{im}\varphi$ and $\operatorname{coim}\varphi$ are isomorphic (which happens in an abelian category) but is strange otherwise. Also, I only know how to prove that $\ker(\operatorname{coker}\varphi)$ satisfies this universal property whenever $\sf{C}$ is abelian. Of course these are nothing but heuristics for me to think that $\ker(\operatorname{coker}\varphi)$ does not satisfy the universal property above.

Answer 1

I think we get a map $\operatorname{coker} (\ker \varphi) \to \ker(\operatorname{coker} \varphi)$, see Tag 0107; for an example where this map is not an isomorphism, see Tag 0108 (in this example, both (the source of) $\ker \varphi$ and (the target of) $\operatorname{coker} \varphi$ are $0$).