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.


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.

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

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.

  • $\begingroup$ 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. $\endgroup$ – user313212 Oct 4 '18 at 11:44
  • $\begingroup$ It's the $f$ that's wrong. The kernel of the zero homomorphism is the identity map $B \to B$. $\endgroup$ – hunter Oct 4 '18 at 12:13
  • $\begingroup$ So do you know what $f$ should I take to prove that part? $\endgroup$ – user313212 Oct 4 '18 at 12:39
  • $\begingroup$ @user313212 yup, take $B \to B/A$ (where I mean quotienting by the image of $A$ of course). $\endgroup$ – hunter Oct 4 '18 at 20:32

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