Let $\mathbf{C}$ be a category with zero object, kernels, and cokernels. Then, a morphism $f\colon A\rightarrow B$ in $\mathbf{C}$ is *semistrict* iff the canonical map $\operatorname{Coker}(\ker (f))\rightarrow \operatorname{Ker}(\operatorname{coker}(f))$ is both a pseudomonomorphism and a pseudoepimorphism (a *pseudobimorphism*). (The term derives from the fact that a preabelian category is said to be (by [some](https://www.math.uni-trier.de/abteilung/analysis/HomAlg.pdf)---the terminology is [not standard](https://ncatlab.org/nlab/show/semi-abelian+category)) *semiabelian* iff this canonical map is always a bimorphism (and of course *abelian* iff this canonical map is always an isomorphism), and the fact that a morphism is said to be *strict* iff this canonical map is an isomorphism.)


Question:



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> Let $\mathbf{C}$ be a finitely-complete finitely-cocomplete category with zero object and let $A\_1,A\_2\in \operatorname{Obj}(\mathbf{C})$. Is it necessarily the case that the projections $\pi \_k\colon A\_1\times A\_2\rightarrow A\_k$ and the inclusions $\iota \_k\colon A\_k\rightarrow A\_1\sqcup A\_2$ are semistrict?
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Disclaimer: I first asked this question on [math.stackexchange](https://math.stackexchange.com/questions/1735513/are-product-coproduct-projections-inclusions-semistrict), and after over a week with no answer, I decided to ask here as well. For what it's worth, while I haven't thought about the question too much since I first asked it, my feeling is that the statement is probably false, simply because (i) my guess is that, were such a statement to be true, its proof would be relatively elementary, and (ii) I spent a couple of days trying all the elementary tricks I could think of (though perhaps I was just being dense . . .).



Are product / coproduct projections / inclusions 'semistrict'?
Here is, I claim, a counterexample. Consider the category $\mathrm{Cat}\_\*$ of pointed categories, i.e. the coslice category $1/\mathrm{Cat}$. Let $A$ be the walking involution, i.e. it has one object $a$ with one nonidentity morphism $e:a\to a$ such that $e e = 1\_a$. Let $B$ be the discrete category on two objects $b,c$, with say $b$ as the basepoint. Then $A\times B$ is just two copies of $A$; consider its projection to $A$. This is surjective, so its cokernel is $1$, and then of course the kernel of the cokernel is $A$.


The kernel of a map in $\mathrm{Cat}\_\*$ is the non-full subcategory of its domain consisting of those objects that map to the basepoint and those morphisms that map to the identity of the basepoint. Thus, the kernel of $\pi:A\times B\to A$ is just $B$. Now the cokernel of this kernel $B\to A\times B$ is the quotient of $A\times B$ obtained by identifying its two objects --- but not doing anything to its morphisms. Thus, it has one object, say $d$, and *two* endomorphisms $e\_1,e\_2:d\to d$ with $e\_1 e\_1 = 1\_d$ and $e\_2 e\_2 = 1\_d$, but no relation between them. In particular, $e\_1 e\_2$ is not the identity, but it does map to the identity in $A$. Hence, the kernel of the induced map $\mathrm{coker}(\mathrm{ker}(\pi)) \to A = \mathrm{ker}(\mathrm{coker}(\pi))$ is nontrivial, so that map is not pseudomonic in your sense, and so $\pi$ is not semistrict.


