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    • CommentRowNumber1.
    • CommentAuthorUrs
    • CommentTimeJan 11th 2015
    • (edited Jan 11th 2015)

    In some topos H, consider some property P on morphisms f:AB (such as being an equivalence, being étale, …), and consider the question of forming “the” (sub-)object [A,B]P of the internal hom [A,B] on those maps satisfying this condition.

    Consider then the special case that H is local with sharp modality . Then a candidate for the internal [A,B]P is the fiber product H(A,B)P×[A,B][A,B].

    (Here H(A,B)PH(A,B) denotes the external space of maps satisfying P, regarded as “codiscretely” embedded back into H.)

    Is that fiber product equivalent to what one ends up with a formulation of [A,B]P in the internal logic?

    • CommentRowNumber2.
    • CommentAuthorMike Shulman
    • CommentTimeJan 12th 2015

    Interesting question; I think the answer is no. A map f:X[A,B] corresponds to a map ˆf:X×AB, hence to a map ˉf:X×AX×B in H/X, and f factors through Equiv(A,B) just when ˉf is an equivalence in H/X. But I think f will factor through your fiber product just when the fiber of ˉf over each point of X is an equivalence, which is a weaker statement.

    • CommentRowNumber3.
    • CommentAuthorUrs
    • CommentTimeJan 12th 2015

    Yes, true, what I wrote sees only the global points. But assuming that A and B and hence also [A,B] are concrete, then it should work, right?

    • CommentRowNumber4.
    • CommentAuthorMike Shulman
    • CommentTimeJan 12th 2015

    Why does concreteness of A and B help? It seems to me like the problem is possible non-concreteness (or maybe non-co-concreteness) of X.

    • CommentRowNumber5.
    • CommentAuthorUrs
    • CommentTimeJan 12th 2015
    • (edited Jan 12th 2015)

    So if A and B are concrete, then so is [A,B] and hence maps into it are characterized by maps into [A,B], which are maps into [A,B] out of (), hence out of all points of the domain.

    • CommentRowNumber6.
    • CommentAuthorMike Shulman
    • CommentTimeJan 12th 2015

    Where by “characterized by” you mean that the map from X[A,B] to X[A,B] is monic, so that two maps X[A,B] are equal as soon as they become so when postcomposing with [A,B][A,B]; hence two maps X×AX×B over X are equal as soon as they become so after pulling back along XX. Right? Why does that imply that a single map X×AX×B over X is an equivalence as soon as it becomes so after pulling back to X? A priori it seems that its inverse X×BX×A might not come from any map over X.

    • CommentRowNumber7.
    • CommentAuthorUrs
    • CommentTimeJan 15th 2015

    I see, thanks!