I found in Borceux' Categorical Algebra the following proposition:
in an abelian category, the following are equivalent:
1) $f:A\longrightarrow B$ is a mono
2) $\operatorname{Ker} f=0$
3) for every $g, f\circ g=0\Rightarrow g=0$
The proof takes several pages and is very complicated, even if not so difficult. I have a very brief proof, and am wondering if it is shorter and correct or simply shorter (but wrong) than Bourceux one. Here is my proof:
1)$\Rightarrow$ 2) Clearly $f\circ 0=0$. If $fg=0$ for some $g$, then $fg=0=f0$, hence $g=0$ since $f$ is mono, hence $g$ factors uniquely through $0$, thus $0$ is a kernel
2)$\Rightarrow$ 3) Let $g$ be such that $fg=0$. Then $g$ factors through $\operatorname{Ker}f=0$, so $g$ is a zero morphism
3)$\Rightarrow$ 1) Assume $f\alpha=f\beta$, for some $\alpha, \beta:C\rightrightarrows A$. Then $$f\alpha-f\beta=0$$ $$f(\alpha -\beta)=0$$ $$\alpha -\beta=0$$ $$\alpha=\beta$$