# Hom functor preserves epimorphisms?

Let $$\cal{C}$$ be a category. Let $$Psh(\cal{C})$$ be the category of presheaves on $$\cal{C}$$. Then there is a functor $$h:\cal{C}\rightarrow Psh(\cal{C})$$ that sends each object $$U$$ in $$\cal{C}$$, a representable presheaf $${\rm{Hom}}_{\cal{C}}(-,U)$$. In other words, $$h$$ is the functor $${\rm{Hom}}_{\cal{C}}(-,-)$$.

Does the functor $$h$$ preserves epimorphisms? More precisely, if $$U\rightarrow V$$ is an epimorphism in $$\cal{C}$$, is the morphism $$h_{U}\rightarrow h_{V}$$ an epimorphism in $$Psh(\cal{C})$$?

Let $$S$$ be a base scheme. If we take $$\cal{C}$$ to be the category $$Sch/S$$ of schemes over $$S$$. Consider the big fppf site $$(Sch/S)_{fppf}$$. Then all representable presheaves on $$(Sch/S)_{fppf}$$ are sheaves. Then is the functor $$h:(Sch/S)_{fppf}\rightarrow Psh((Sch/S)_{fppf})$$ preserves epimorphisms? More precisely, if $$U\rightarrow V$$ is an epimorphism in $$Sch$$, is the morphism $$h_{U}\rightarrow h_{V}$$ an epimorphism in the category of sheaves on $$(Sch/S)_{fppf}$$?

Since for each scheme $$U$$, the representable sheaf $$h_{U}$$ is an algebraic space. And it has been proved that if if $$U\rightarrow V$$ is a monomorphism in $$Sch$$, then the morphism $$h_{U}\rightarrow h_{V}$$ is a monomorphism in the category of algebraic spaces.

• Do you have reason to suspect this is true? Is this part of a larger problem? Have you any thoughts so far.. ? Commented Oct 25, 2022 at 13:04
• @FShrike: I will give a more concrete thought later. Commented Oct 25, 2022 at 13:06

$$\newcommand{\D}{\mathfrak{D}}\newcommand{\set}{\mathsf{Set}}\newcommand{\psh}{\operatorname{Psh}}\newcommand{\H}{\mathsf{H}}$$I will use my own notation out of preference, but everything I say here is easily translated to your notation.

Let $$\D$$ be a locally small category, and $$\H:\D\to\psh_\D$$ the Yoneda embedding. I claim that an arrow $$U\overset{\alpha}{\longrightarrow}V$$ in $$\D$$ has $$\H(\alpha):\H(U)\implies\H(V)$$ an epimorphism in $$\psh_\D$$ if and only if $$\alpha$$ is a split epimorphism.

Exercise $$6.2.20$$ in Leinster's basic category theory ask one to show that, if the codomain has all pullbacks, arrows in functor categories (=natural transformations) are mono/epi iff. every single one of their components is. I will give or sketch my solution to this exercise if you wish. Suffice it to say, $$\H(\alpha)$$ is an epimorphism iff. every component is.

Note: Leinster phrases the result for small domains and locally small codomains only. In private communication with him, he said that most of his insistence on small categories was to avoid foundational quibbles in his book, which was to be as simple as possible. However, if you don't mind the possibility that $$\psh_\D$$ is not locally small, then don't worry!

That is to say, if and only if every $$\D(A,U)\overset{\alpha\,\circ\,-}{\longrightarrow}\D(A,V)$$ is an epimorphism in $$\set$$, i.e. a surjection, for all $$A\in\D$$. That is true if and only if every arrow $$A\to V$$ factors via $$A\longrightarrow U\overset{\alpha}{\longrightarrow}V$$.

Suppose $$\alpha$$ is a split epimorphism: by definition, that is iff. there is a 'section' $$\beta:V\to U$$ that $$\alpha\circ\beta=1_V$$. Then given $$f:A\to V$$, I can define $$g=\beta\circ f:A\to U$$: $$\alpha\circ g=f$$ is a desired factorisation.

Conversely, suppose $$\alpha$$ has this property for all $$A$$. I can then specify $$A=V$$, and say that the identity $$1_V$$ factors through $$\alpha$$. That means there is $$\beta:V\to U$$ that $$\alpha\circ\beta=1_V$$, i.e. $$\alpha$$ is a split epimorphism.

In conclusion, $$\H(\alpha)$$ is an epimorphism of $$\psh_\D$$ iff. $$\alpha$$ is a split epimorphism. A little whimsically, if your category satisfies the axiom of choice, then $$\H$$ indeed preserves epimorphisms.

• What confuses me is that the category $Sch$ of schemes defaultly satisfies the axiom of choice. But it is not obvious that every epimorphism in $Sch$ splits. Commented Oct 25, 2022 at 13:45
• @ToneyLeung Oh, to me the "axiom of choice" for a category is precisely the statement: "every epimorphism splits" (by analogy with $\mathsf{Set}$). What does it mean for you? Commented Oct 25, 2022 at 13:49
• We defaultly use ZFC to avoid proper classes. Commented Oct 25, 2022 at 13:51
• @ToneyLeung Oh I see, you just mean the normal axiom of choice holds. That wasn't what I meant by the axiom of choice for a category, since the arrows need not be functions between sets. I don't anything about schemes, so you'll have to decide for yourself which epimorphisms in $\mathsf{Sch}$ split. That might be worth asking in a separate MSE post Commented Oct 25, 2022 at 13:53