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Let $\mathsf{C}$ be a category with initial and terminal objects, and $\phi:\mathscr{F}\to\mathscr{G}$ a morphism of presheaves on $X$ taking values in $\mathsf{C}$. I have a rather messy proof that if $\phi$ is a monomorphism (resp. epimorphism), then so is $\phi(U):\mathscr{F}(U)\to\mathscr{G}(U)$ for every open subset $U\subset X$. I'm wondering if there is any elegant proof?

Mike Pierce
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ashpool
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1 Answers1

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Assume $\phi$ is a monomorphism. Assume $\phi(U)\circ f=\phi(U)\circ g$ with $f,g\colon \mathscr G(U)\to A$. Consider presheaf $\mathscr A$ with $\mathscr A(V)=A$ if $U\subseteq V$, $\mathscr A(V)=0$ otherwise (with obvious restrictions). Then $f$ allows us to define $\psi_f\colon\mathscr G\to \mathscr A$ by letting $\psi_f(V)\colon \mathscr G(V)\to \mathscr A(V)$ be the morphism $f\circ \operatorname{res}^V_{U}$ if $U\subseteq V$ (and $0$ otherwise). Similarly, we find $\psi_g$. As $\phi$ is a monomorphism, $\psi_f=\psi_g$, especially, $f=\psi_f(U)=\psi_g(U)=g$, hence $\phi(U)$ is a monomorphism.

Can you see how to dualize this to handle epimorphisms?

Hagen von Eitzen
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  • Yes, my "messy" proof is along the line you suggested. But I was wondering if I can make use of more fundamental properties from category theory, like adjoints. – ashpool Feb 10 '13 at 20:18
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    You cannot really have a proof easier than this one. I think it would be overkill to try to use other machinery than just the definitions. – Bogdan Feb 10 '13 at 21:21
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    @ashpool For a more "categoric" proof see the accepted and bountified answer by Akhil Mathew here. – Hagen von Eitzen Feb 10 '13 at 22:26
  • Hint: Right adjoints preserve limits, in particular monomorphisms. – Martin Brandenburg Feb 12 '13 at 15:30
  • @Hagen von Eitzen: Sry for commenting this old post. But I think you mixed up the proof for mono- and epimorphisms. For example f,g should be arrows from A -> F(U). – felix Jan 03 '21 at 19:49