# 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?

• The kernel of the cokernel and the cokernel of the kernel just don't compute the right thing unless you're in an $\text{Ab}$-enriched category (where they compute two variations on the image). What examples are you interested in that aren't $\text{Ab}$-enriched? – Qiaochu Yuan Apr 10 '16 at 7:38
• $\mathbf{Grp}$ is perhaps the most natural nonadditive category I had in mind. – Jonathan Gleason Apr 10 '16 at 7:41
• Not an answer to your question, but in any homological (i.e. regular pointed protomodular) category, such as the category of groups, the cokernel of the kernel is still the image. Therefore in such a category any regular epimorphism (such as a projection morphism) is not just semistrict but strict. Generally, in any pointed category, the comparison map in the case of a projection $A_1\times A_2\xrightarrow{\pi_1}A_1$ is a split epimorphism $Coker(i_2)\to A_1$ for $A_2\xrightarrow{i_2}A_1\times A_2$ the natural inclusion. If there is a counter-example, I expect it to be a monoid. – Vladimir Sotirov Apr 10 '16 at 21:08