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    • CommentRowNumber1.
    • CommentAuthoradeelkh
    • CommentTimeApr 4th 2015
    • (edited Apr 5th 2015)

    Let CC be an \infty-site and let φ:FG\varphi : F \to G be a morphism of presheaves on CC. I believe it is true that the morphism of associated sheaves is an effective epimorphism iff φ\varphi is “section-wise locally surjective”, i.e. for any section sπ 0G(c)s \in \pi_0 G(c) there exists a covering (c αc)(c_\alpha \to c) such that the restrictions s|c αs|c_\alpha lift to sections s α˜π 0F(c α)\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 FF and GG are 0-truncated, then φ\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 φ\varphi is automatically 0-truncated). This is easy to check because I can use the above section-wise condition.

    If FF and GG 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 φ\varphi being “section-wise locally injective”?

    • CommentRowNumber2.
    • CommentAuthorMike Shulman
    • CommentTimeApr 4th 2015

    This sounds like it would be a fun exercise to work out in the internal HoTT.

    • CommentRowNumber3.
    • CommentAuthorZhen Lin
    • CommentTimeApr 4th 2015

    The only simple condition I can think of is that ϕ:FG\phi : F \to G is already a monomorphism of presheaves. Of course, this is only sufficient and not necessary.

    • CommentRowNumber4.
    • CommentAuthorMike Shulman
    • CommentTimeApr 4th 2015

    With the usual indexing, being a monomorphism means being (1)(-1)-truncated. Is the question about (1)(-1)-truncatedness or 0-truncatedness?

    • CommentRowNumber5.
    • CommentAuthorZhen Lin
    • CommentTimeApr 4th 2015

    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.

    • CommentRowNumber6.
    • CommentAuthoradeelkh
    • CommentTimeApr 4th 2015
    • (edited Apr 4th 2015)

    In order to prove φ\varphi is a local isomorphism, I only need 0-truncatedness, given that φ\varphi is 0-connected. This is equivalent to the diagonal of φ\varphi being (-1)-truncated, so a sufficient condition for that would already be useful.

    (Of course, I’m assuming that CC admits the relevant limits.)

    • CommentRowNumber7.
    • CommentAuthorMarc Hoyois
    • CommentTimeApr 4th 2015

    If you know that the sheafification of ϕ\phi is hypercomplete, then it’s nn-truncated iff ϕ\phi has the local right lifting property with respect to S k*S^k \to * for kn+1k\geq n+1. In general I don’t see how to characterize nn-truncatedness in such a way.

    • CommentRowNumber8.
    • CommentAuthoradeelkh
    • CommentTimeApr 4th 2015

    Marc: hypercompletion for a morphism means that it is a local equivalence with respect to the class of infinity-connected morphisms?

    • CommentRowNumber9.
    • CommentAuthorMike Shulman
    • CommentTimeApr 4th 2015

    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?

    • CommentRowNumber10.
    • CommentAuthorMarc Hoyois
    • CommentTimeApr 4th 2015

    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.

    • CommentRowNumber11.
    • CommentAuthorMarc Hoyois
    • CommentTimeApr 4th 2015

    In general nn-truncatedness is equivalent to the RLP with respect to nn-connected morphisms of presheaves, but that’s not very explicit.

    • CommentRowNumber12.
    • CommentAuthoradeelkh
    • CommentTimeApr 4th 2015

    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 φ\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).

    • CommentRowNumber13.
    • CommentAuthorMike Shulman
    • CommentTimeApr 5th 2015
    • (edited Apr 5th 2015)

    I see. Any particular reason you chose n=0n=0? It’s equally true that you could prove its sheafification is nn-connected using a section-wise condition and then wonder about nn-truncatedness, for any n2n\ge -2.

    • CommentRowNumber14.
    • CommentAuthorMike Shulman
    • CommentTimeApr 5th 2015

    The only fully general way I can think of to show that a map ff has an nn-truncated sheafification is to show that the (n+2)(n+2)-fold iterated diagonal of ff itself is a local equivalence. But that probably doesn’t help you a whole lot.

    • CommentRowNumber15.
    • CommentAuthoradeelkh
    • CommentTimeApr 5th 2015

    Sure, that is true.

    • CommentRowNumber16.
    • CommentAuthoradeelkh
    • CommentTimeApr 5th 2015
    • (edited Apr 5th 2015)

    Now I’ve managed to confuse myself on something really silly. Let φ:FG\varphi : F \to G be an arbitrary morphism of presheaves and consider the diagonal Δ:FF× GF\Delta : F \to F \times_G F. Δ\Delta is a monomorphism iff Δ(c):F(c)F(c)× G(c)F(c)\Delta(c) : F(c) \to F(c) \times_{G(c)} F(c) is a monomorphism of infinity-groupoids for each object cCc \in C. Let cCc \in C be an object and let (x,y):*F(c)× G(c)F(c)(x,y) : * \to F(c) \times_{G(c)} F(c) be a point in the target, corresponding to points x:*F(c)x : * \to F(c) and y:*F(c)y : * \to F(c) such that φ(c)(x)=φ(c)(y)\varphi(c)(x) = \varphi(c)(y). If the fibre at (x,y)(x,y) is nonempty, let zz be a point in it, which is by definition a point of F(c)F(c) such that Δ(c)(z)=(z,z)=(x,y)\Delta(c)(z) = (z,z) = (x,y). Then it is clear that the point zz is unique, hence every fibre of Δ(c)\Delta(c) is empty or contractible, hence Δ(c)\Delta(c) is a monomorphism for all cc, hence Δ\Delta is a monomorphism.

    Where am I implicitly using that F(c)F(c) and G(c)G(c) are discrete here?

    • CommentRowNumber17.
    • CommentAuthorMarc Hoyois
    • CommentTimeApr 6th 2015

    The point zz may not be unique. For instance, take ϕ:BG*\phi: B G \to * for GG a discrete group (with C=*C=*, say). The diagonal is BGB(G×G)B G \to B(G\times G). The fiber over the base point is the set of orbits (G×G)/G(G\times G)/G.

    • CommentRowNumber18.
    • CommentAuthoradeelkh
    • CommentTimeApr 6th 2015

    Ah right, thanks.