1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Doesn't talking about kernels in a category presuppose having a zero object in that category? $\endgroup$ Commented Jun 9, 2020 at 3:03
  • $\begingroup$ 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)? $\endgroup$
    – CJHJ
    Commented Jun 9, 2020 at 4:01

1 Answer 1

1
$\begingroup$

EDIT: As pointed out in the comments by @Geoffrey Trang, the same argument applies to monomorphisms and dually to epimorphisms.

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 $0$.

$\endgroup$
3
  • $\begingroup$ 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. $\endgroup$ Commented Jun 9, 2020 at 4:41
  • $\begingroup$ Good point, I'll edit to mention this. $\endgroup$ Commented Jun 9, 2020 at 4:44
  • $\begingroup$ Thanks! I forgot the fact that if a zero object exists in a category, than it should be isomorphic to the terminal object. Everything makes sense now. $\endgroup$
    – CJHJ
    Commented Jun 9, 2020 at 4:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .