Surjective maps from functors of points to presheaves Let $\mathcal{C}$ be a category. Let $F:\mathcal{C}^{opp}\rightarrow Sets$ be a presheaves of sets. Is it possible to find a functor of points $h_{U}$ for $U\in{\rm{Ob}}(\mathcal{C})$ such that $h_{U}\rightarrow F$ is surjective as a map of presheaves? More concretely, if $\mathcal{C}=Sch/S$ where $S$ is a base scheme, is it possible to find some $U\in{\rm{Ob}}(Sch/S)$ such that $h_{U}\rightarrow F$ is surjective on sections? It seems that it is not always possible.
 A: *

*To be an epimorphism $X \to Y$ of presheaves, each component map $X (U) \to Y (U)$ must be surjective.


*In particular, if there is a $U$ such that $Y (U)$ has strictly greater cardinality than $X (U)$, then there can be no epimorphism $X \to Y$.


*If $\mathcal{C}$ has at least one object – say $U$ – then we can construct a presheaf $Y$ such that $Y (U)$ has any cardinality we like: take the constant presheaf!
If you don't like the above argument, here is another:

*

*Representable presheaves are connected, i.e. if $X$ is representable then for any family $Y_i$ ($i \in I$) of presheaves the canonical comparison map
$$\coprod_{i \in I} \textrm{Hom} (X, Y_i) \longrightarrow \textrm{Hom} \left( X, \textstyle \coprod_{i \in I} Y_i \right)$$
is a bijection.
That means every morphism $X \to \coprod_{i \in I} Y_i$ must factor through exactly one of the coproduct cocone components $Y_j \to \coprod_{i \in I} Y_i$.


*Hence, for any object $U$ in $\mathcal{C}$, if $Y$ and $Z$ are presheaves such that $Y (U)$ and $Z (U)$ are non-empty, then no morphism $h_U \to Y \amalg Z$ can be an epimorphism.
A: A presheaf with no elements, i.e. the constant presheaf $F$ with $F(U)=\emptyset$, is a strict initial object: the only natural transformations to it are isomorphisms, hence also come from presheaves with no elements. But representable presheaves always have elements: $h_U(U)$ always contains the identity morphism. Thus there are no natural transformations $h_U\Rightarrow F$ from a representable functor to the empty presheaf.
More generally, a necessary condition for a presheaf $F$ to admit a surjective natural transformation from a representable $h_U$ is that $F$ be connected: that the natural map $\mathrm{Hom}(F,G)\sqcup\mathrm{Hom}(F,H)\to\mathrm{Hom}(F,G\sqcup H)$ be an isomorphism. This is indeed more general, as
the empty presheaf is not connected: $\mathrm{Hom}(F,F)\sqcup\mathrm{Hom}(F,F)\to\mathrm{Hom}(F,F\sqcup F)=\mathrm{Hom}(F,F)$ is not an isomorphism since it sends a set with two elements to a set with a single element.
The reason for the necessary condition is that on the one hand, representables $h_U$ are connected: $\mathrm{Hom}(h_U,G)\sqcup\mathrm{Hom}(h_U,H)\cong G(U)\sqcup H(U)\cong\mathrm{Hom}(h_U,G\sqcup H)$. On the other hand, an epimorphism (e.g. surjective natural transformation) whose domain is connected has to have its codomain connected as well (see https://ncatlab.org/nlab/show/connected+object).
