Does the kernel of a morphism equal to its source object in an abelian category imply it is a zero map?

Question

Does the kernel of a morphism equals to its source object in abelian category implies this is a zero map?

Namely let $C$ be an abelian category, $X,Y$ are two objects, $f\in Hom_{C}(X,Y)$ is a morphism, then if $\ker(f)=X$, can we conclude $f$ is the zero map?

Answer 1

Assuming that you mean that the identity map $1_X:X\to X$ is the kernel of $f$, then yes. By definition, if $k:K\to X$ is the kernel of $f$, $fk=0$ (and $k$ is universal with this property). So if $1_X$ is the kernel, then $0=f1_X=f$.

If you just mean that there is some morphism $g:X\to X$ which is a kernel of $f$, then no. For instance, let $f:\mathbb{Z}\to\mathbb{Z}/2$ be the quotient map in $Ab$ and leg $g:\mathbb{Z}\to\mathbb{Z}$ be multiplication by $2$. Then $g$ is a kernel of $f$, but $f\neq 0$.