Kernels in additive categories

Question

I have a problem regarding kernels. I know the definition of a kernel as an equalizer. My problem is the following, say we have a morphism $f: X \rightarrow Y$ such that the kernel is isomorphic to $X$. Now the kernel is a pair consisting of an object and a morphism. Does this now mean that the morphism $\ker(f) \rightarrow X$ is an isomorphism or does it simply mean that there is an isomorphism from the kernel to $X$. In the latter case, how does one proof that f is the zero morphism? As is said in the title, the category is an additive category. Thanks in advance.

Answer 1

Let's start with the case where kernel morphism $k$ is an isomorphism. Since it by definition commute with equalising arrows, we have $f \circ k = 0 \circ k = 0$. Composing both sides with its inverse results in $$ f = f \circ k \circ k^{-1} = 0 \circ k^{-1} = 0 $$

In the second case the statement is not true. Consider for example the projection onto the first component from the countable product in the category of abelian groups $$ \prod^\infty \mathbb Z \xrightarrow{p} \mathbb Z $$ Clearly the kernel of $p$ is isomorphic to the domain, but $f$ is not $0$.