0

Consider $T$ to be a terminal element, $T \stackrel{f}{\rightarrowtail} A$ be a monic morphism (this can be shown by the terminal property of $T$) and $B\stackrel{\tau_B}{\rightarrow} T$ be the unique terminal morphism from $B$ to $T$. Then does this make $B \stackrel{f \circ \tau_B}{\rightarrow} A$ a monomorphism? Why?

Felipe Dilho
  • 207
  • 1
    Have you thought about any examples (in $\bf{Set}$ for instance)? – S.C. Jan 30 '24 at 05:37
  • Actually this comes from another question that I had, in which there where more structure but I thought that not to be inportant for this inference, apparently it is: https://math.stackexchange.com/questions/4853463/how-do-i-prove-the-unicity-of-the-terminal-morphism-obtained-by-being-a-equivale – Felipe Dilho Jan 30 '24 at 05:46
  • In the answer there it is said from the outset that for those $g,h$, $g \circ h$ was monic, and I could not see why, after thinking about it. – Felipe Dilho Jan 30 '24 at 05:47

1 Answers1

1

In the category of R-modules, $0$ is a terminal object. The morphism $R \longrightarrow 0 \longrightarrow R$ is not monic.

Functor
  • 1,661