This is part of Weibel's Exercise 1.2.2, where I have to show that in the category R-Mod, every monic is a kernel.
A monic morphism is defined to be a map $i \colon A \to B$ such that if $g \colon A \to A$ is such that $i \circ g = 0$, then $g = 0$, and a kernel of a homomorphism $f \colon B \to C$ is a map $i \colon A \to B$ such that $f \circ i = 0$ and $i$ is universal.
Attempt
Take a monic map $i \colon A \to B$ and $f \colon B \to 0$ be the zero homomorphism. Then obviously $f \circ i = 0$ and also if $i' \colon A \to B$ is such that $f \circ i' = 0$ there is a unique map $u \colon A' \to A$ defined to be the composition $$u \colon A' \to 0 \to A$$ It is unique since it is the composition of the zero map $A' \to 0$ and the zero map $0 \to A$.
It seems to be a problem with my attempt since I didn't use the fact that $i$ is monic, but I can't find any flaws in the argument, so I'd like if anyone could point them out and maybe a hint to complete the exercise. Thank you.
The map $u: A'\to A$ is not unique such that the composition $A' \to A \to B$ is zero. Indeed, any map $A'\to A$ will have the property that the composition is zero.