Schur's lemma in abelian categories

Remarks

The kernel of $f$ is a subobject of $S$, so it is $0$ or $S$. The image of $f$ is a subobject of $T$, so it is $0$ or $T$. If $f \ne 0$, then the kernel is not $S$ and the image is not $0$. So $f$ is a monomorphism and an epimorphism, and in an abelian category such a morphism is an isomorphism. For $\End(S)$, every nonzero element is invertible, and $1 \ne 0$ because $S$ is not zero. In the category of modules over a ring, the simple objects are the simple modules, and one gets the classical Schur's lemma. The simplicity hypothesis is needed: multiplication by $2$ on the $\Z$-module $\Z$ is nonzero and not an isomorphism, and $\Z$ is not simple since $2\Z$ is a proper nonzero submodule.

Going deeper

For a student

The lemma is the first step in studying objects of finite length, which are built from simple objects by extensions as in the Jordan–Hölder theorem. It shows that $\Hom(S, T) = 0$ for non-isomorphic simple objects and that $\End(S)$ is a division ring. In representation theory over an algebraically closed field $k$, a further argument with eigenvalues gives $\End(S) = k$ for finite-dimensional simple representations; that sharper form needs the extra hypotheses.