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Let $C$ be an $\infty$-site and let $\varphi : F \to G$ be a morphism of presheaves on $C$. I believe it is true that the morphism of associated sheaves is an effective epimorphism iff $\varphi$ is “section-wise locally surjective”, i.e. for any section $s \in \pi_0 G(c)$ there exists a covering $(c_\alpha \to c)$ such that the restrictions $s|c_\alpha$ lift to sections $\tilde{s_\alpha} \in \pi_0 F(c_\alpha)$. (Does anyone know the reference in HTT for this, by the way?) (Edit: it follows from 7.2.1.14 in HTT.)
If $F$ and $G$ are 0-truncated, then $\varphi$ is a local isomorphism iff the morphism of associated sheaves is 0-connected, i.e. it and its diagonal are effective epimorphisms (because then $\varphi$ is automatically 0-truncated). This is easy to check because I can use the above section-wise condition.
If $F$ and $G$ are not 0-truncated, is there a simple condition that can be checked to ensure that the morphism of associated sheaves is 0-truncated, something along the lines of the diagonal of $\varphi$ being “section-wise locally injective”?
This sounds like it would be a fun exercise to work out in the internal HoTT.
The only simple condition I can think of is that $\phi : F \to G$ is already a monomorphism of presheaves. Of course, this is only sufficient and not necessary.
With the usual indexing, being a monomorphism means being $(-1)$-truncated. Is the question about $(-1)$-truncatedness or 0-truncatedness?
It seems to me that, in this situation, 0-truncatedness implies (-1)-truncatedness. The point being that, for 0-truncated morphisms, the (relative) diagonal is automatically a (-1)-truncated, hence is an isomorphism if and only if it is an effective epimorphism.
In order to prove $\varphi$ is a local isomorphism, I only need 0-truncatedness, given that $\varphi$ is 0-connected. This is equivalent to the diagonal of $\varphi$ being (-1)-truncated, so a sufficient condition for that would already be useful.
(Of course, I’m assuming that $C$ admits the relevant limits.)
If you know that the sheafification of $\phi$ is hypercomplete, then it’s $n$-truncated iff $\phi$ has the local right lifting property with respect to $S^k \to *$ for $k\geq n+1$. In general I don’t see how to characterize $n$-truncatedness in such a way.
Marc: hypercompletion for a morphism means that it is a local equivalence with respect to the class of infinity-connected morphisms?
Can you state the exact situation and the question again? I am very confused. What are you assuming about the morphism and what do you want to prove?
Adeel: No, I mean it’s a hypercomplete object in the slice topos. In that case n-truncatedness can be checked by the vanishing of homotopy sheaves (in the slice topos), which I think amounts to what I wrote.
In general $n$-truncatedness is equivalent to the RLP with respect to $n$-connected morphisms of presheaves, but that’s not very explicit.
Mike: basically, I have a morphism of presheaves on an $\infty$-site which I want to prove is an isomorphism after sheafification. I can prove its sheafification is 0-connected by using the section-wise condition I mentioned. So it remains to show that its sheafification is 0-truncated, and I was wondering whether there was a section-wise condition on $\varphi$ that I could use to check this.
Marc: I see, thanks. I’m not sure at the moment whether this applies in my situation, I’ll have to check (I don’t have a good intuition for hypercomplete objects).
I see. Any particular reason you chose $n=0$? It’s equally true that you could prove its sheafification is $n$-connected using a section-wise condition and then wonder about $n$-truncatedness, for any $n\ge -2$.
The only fully general way I can think of to show that a map $f$ has an $n$-truncated sheafification is to show that the $(n+2)$-fold iterated diagonal of $f$ itself is a local equivalence. But that probably doesn’t help you a whole lot.
Sure, that is true.
Now I’ve managed to confuse myself on something really silly. Let $\varphi : F \to G$ be an arbitrary morphism of presheaves and consider the diagonal $\Delta : F \to F \times_G F$. $\Delta$ is a monomorphism iff $\Delta(c) : F(c) \to F(c) \times_{G(c)} F(c)$ is a monomorphism of infinity-groupoids for each object $c \in C$. Let $c \in C$ be an object and let $(x,y) : * \to F(c) \times_{G(c)} F(c)$ be a point in the target, corresponding to points $x : * \to F(c)$ and $y : * \to F(c)$ such that $\varphi(c)(x) = \varphi(c)(y)$. If the fibre at $(x,y)$ is nonempty, let $z$ be a point in it, which is by definition a point of $F(c)$ such that $\Delta(c)(z) = (z,z) = (x,y)$. Then it is clear that the point $z$ is unique, hence every fibre of $\Delta(c)$ is empty or contractible, hence $\Delta(c)$ is a monomorphism for all $c$, hence $\Delta$ is a monomorphism.
Where am I implicitly using that $F(c)$ and $G(c)$ are discrete here?
The point $z$ may not be unique. For instance, take $\phi: B G \to *$ for $G$ a discrete group (with $C=*$, say). The diagonal is $B G \to B(G\times G)$. The fiber over the base point is the set of orbits $(G\times G)/G$.
Ah right, thanks.
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