# Every monic is a kernel

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.

• Typo in definition of kernel? (Should $i$ map $C \to A$?) – hunter Oct 4 '18 at 11:31
• @hunter or in the definition of $f$ ($f \colon B \to C$). I'll edit it now. – user313212 Oct 4 '18 at 11:33
• Try $f: B \to B/i(A)$ – leibnewtz Oct 4 '18 at 12:48

## 1 Answer

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.

• Okay, I think I get it know. So the problem is that the $u$ I chose is not the one that works, or is it the $f$ I took? Thank you. – user313212 Oct 4 '18 at 11:44
• It's the $f$ that's wrong. The kernel of the zero homomorphism is the identity map $B \to B$. – hunter Oct 4 '18 at 12:13
• So do you know what $f$ should I take to prove that part? – user313212 Oct 4 '18 at 12:39
• @user313212 yup, take $B \to B/A$ (where I mean quotienting by the image of $A$ of course). – hunter Oct 4 '18 at 20:32