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

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.

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 $$0$$).