Let $\mathcal{C}$ be a site and $f:\mathcal{F}\to \mathcal{G}$ a morphism of $2$-sheaves. According to https://mathoverflow.net/q/307366, this is an epimorphism if and only if it is almost surjective, that is to say if for all $U$, and any $g\in G(U)$, there exists an open cover $\{U_i\to U\}$ such that there exists $x_i\in F(U_i)$ such that $f(x_i)\cong g\rvert_{U_i}$. If we assume that our $2$-topos has enough points, can we check this on stalks, that is to say on the pullback of $f$ to all points of our $2$-topos?
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2$\begingroup$ What exactly do you mean by an "$n$-sheaf"? In the most common higher-categorical formulation, an $n$-sheaf lives in a weak $(n+1)$-category, where it doesn't even make sense to ask about whether something is an isomorphism (versus an equivalence). $\endgroup$– Mike ShulmanCommented Feb 21, 2021 at 2:21
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$\begingroup$ Oh, right, how embarassing! Apologies, I'll change the question accordingly, since I'm still curious about the essential surjectivity. $\endgroup$– curious math guyCommented Feb 21, 2021 at 9:55
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1$\begingroup$ A conservative functor reflects anything (not really – but I don't want to be precise here) it preserves. The inverse image functor of a geometric morphism preserves finite limits and (possibly infinite) colimits, so in particular it preserves images, and hence reflects the property of being essentially surjective. But the functor $F \mapsto F (U)$ is not usually the inverse image functor of a geometric morphism. You seem to be confusing "sections" vs "stalks". $\endgroup$– Zhen LinCommented Feb 21, 2021 at 12:30
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2$\begingroup$ Oh, I missed that your previous comment was about the image sheaf surjecting section-wise, unlike the original question. That's true for 1-sheaves and also higher sheaves. In fact, since the image is a subsheaf, it surjects on stalks iff it surjects sectionwise iff it is an isomorphism. $\endgroup$– Mike ShulmanCommented Feb 26, 2021 at 2:25
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1$\begingroup$ I'm not necessarily saying you were wrong to change the question. This version doesn't look to me like the same question you started with, since it says nothing about surjectivity on sections, but I can appreciate that this is probably the question you meant to ask. $\endgroup$– Mike ShulmanCommented Feb 26, 2021 at 17:07
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By definition (e.g. Remark 6.5.4.7 of Higher topos theory), an $n$-topos $\mathcal{E}$ has enough points if for every morphism $f:X\to Y$ in $\mathcal{E}$, whenever $p^*(f)$ is an equivalence for all points $p:\mathcal{S}_n \to \mathcal{E}$ (where $\mathcal{S}_n$ is the $n$-topos of $(n-1)$-groupoids), then $f$ is already an equivalence.
What you call an "epimorphism" or "almost surjective" is perhaps better called an effective epimorphism, since it is a faithful generalization of that concept from 1-categories, whereas it is not really a faithful generalization of the concept of epimorphism (e.g. it is not the same as a monomorphism in the opposite $n$-category). By Prop. 7.2.1.14 of Higher topos theory, a morphism in an $n$-topos is an effective epimorphism precisely when its truncation to a morphism of 1-sheaves is an (effective) epimorphism in the corresponding 1-topos; this shows that the effective epimorphisms agree with your concrete description.
Now the effective epimorphisms in an $n$-topos are the left class of a factorization system whose right class are the monomorphisms. Thus, a morphism is an effective epimorphism precisely if the monomorphism half of this factorization is an equivalence. It follows that being an effective epimorphism is reflected by any conservative family of functors, such as the family of points of an $n$-topos with enough points.
answered Feb 26, 2021 at 17:06