# Is terminal object the kernel of identity morphism?

Let's say that there is a category $$\mathbf{C}$$ with $$A$$ being an object of that category and a zero object exists in that category. If we have an identity morphism $${id}_A: A\to A$$, is the kernel of this morphism a terminal object of the category? My reasoning for this stems from the fact that the universal property of kernel requires a unique morphism going into $$\ker(id)$$, hence the terminal object.

• Doesn't talking about kernels in a category presuppose having a zero object in that category? Commented Jun 9, 2020 at 3:03
• Ah, yes it does. Edited the question now. Does that mean the kernel of the mapping should be the zero object (zero object being one of the terminal object)?
– CJHJ
Commented Jun 9, 2020 at 4:01

In fact, you can show that any isomorphism $$f : A \longrightarrow B$$ satisfies $$ker(f) = 0$$. Indeed, let $$g: C \longrightarrow A$$ such that $$f \circ g = 0$$. We want to show that $$g$$ factors through $$0$$. Well since $$f$$ is an isomorphism, $$g = 0$$. Hence, $$g$$ factors as $$C \longrightarrow 0 \longrightarrow A$$ and we have shown that $$0$$ satisfies the universal propertt of the kernel. Dually, the cokernel of an isomorphism is $$0$$.
• Even more generally, monomorphisms and epimorphisms have zero kernel and cokernel respectively. The converse is true in preadditive categories ($\mathbf{Ab}$-enriched categories), but not in general. Commented Jun 9, 2020 at 4:41