Are product / coproduct projections / inclusions 'semistrict'? 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---the terminology is not standard) 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:

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?

Disclaimer:  I first asked this question on math.stackexchange, 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 . . .).
 A: 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.
