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1. • CommentRowNumber1.
   • CommentAuthorUrs
   • CommentTimeJan 11th 2015
   • (edited Jan 11th 2015)
   • PermaLink
   Author: Urs
   Format: MarkdownItexIn some topos $\mathbf{H}$, consider some property $P$ on morphisms $f \colon A \to B$ (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 $\mathbf{H}$ is [[local topos|local]] with [[sharp modality]] $\sharp$. Then a candidate for the internal $[A,B]_P$ is the fiber product $\mathbf{H}(A,B)_P\underset{\sharp [A,B]}{\times} [A,B]$. (Here $\mathbf{H}(A,B)_P \hookrightarrow \mathbf{H}(A,B)$ denotes the external space of maps satisfying $P$, regarded as "codiscretely" embedded back into $\mathbf{H}$.) Is that fiber product equivalent to what one ends up with a formulation of $[A,B]_P$ in the internal logic?

   In some topos $\mathbf{H}$, consider some property $P$ on morphisms $f \colon A \to B$ (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 $\mathbf{H}$ is local with sharp modality $\sharp$. Then a candidate for the internal $[A,B]_P$ is the fiber product $\mathbf{H}(A,B)_P\underset{\sharp [A,B]}{\times} [A,B]$.

   (Here $\mathbf{H}(A,B)_P \hookrightarrow \mathbf{H}(A,B)$ denotes the external space of maps satisfying $P$, regarded as “codiscretely” embedded back into $\mathbf{H}$.)

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

2. • CommentRowNumber2.
   • CommentAuthorMike Shulman
   • CommentTimeJan 12th 2015
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   Author: Mike Shulman
   Format: MarkdownItexInteresting question; I think the answer is no. A map $f:X\to [A,B]$ corresponds to a map $\hat{f} : X\times A \to B$, hence to a map $\bar{f}:X\times A \to X\times B$ in $\mathbf{H}/X$, and $f$ factors through $Equiv(A,B)$ just when $\bar{f}$ is an equivalence in $\mathbf{H}/X$. But I think $f$ will factor through your fiber product just when the fiber of $\bar{f}$ over each point of $X$ is an equivalence, which is a weaker statement.

   Interesting question; I think the answer is no. A map $f:X\to [A,B]$ corresponds to a map $\hat{f} : X\times A \to B$, hence to a map $\bar{f}:X\times A \to X\times B$ in $\mathbf{H}/X$, and $f$ factors through $Equiv(A,B)$ just when $\bar{f}$ is an equivalence in $\mathbf{H}/X$. But I think $f$ will factor through your fiber product just when the fiber of $\bar{f}$ over each point of $X$ is an equivalence, which is a weaker statement.

3. • CommentRowNumber3.
   • CommentAuthorUrs
   • CommentTimeJan 12th 2015
   • PermaLink
   Author: Urs
   Format: MarkdownItexYes, 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?

   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?

4. • CommentRowNumber4.
   • CommentAuthorMike Shulman
   • CommentTimeJan 12th 2015
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexWhy does concreteness of $A$ and $B$ help? It seems to me like the problem is possible non-concreteness (or maybe non-co-concreteness) of $X$.

   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 $X$.

5. • CommentRowNumber5.
   • CommentAuthorUrs
   • CommentTimeJan 12th 2015
   • (edited Jan 12th 2015)
   • PermaLink
   Author: Urs
   Format: MarkdownItexSo if $A$ and $B$ are concrete, then so is $[A,B]$ and hence maps into it are characterized by maps into $\sharp [A,B]$, which are maps into $[A,B]$ out of $\flat(-)$, hence out of all points of the domain.

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

6. • CommentRowNumber6.
   • CommentAuthorMike Shulman
   • CommentTimeJan 12th 2015
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexWhere by "characterized by" you mean that the map from $X\to [A,B]$ to $X\to \sharp [A,B]$ is monic, so that two maps $X\to [A,B]$ are equal as soon as they become so when postcomposing with $[A,B]\to \sharp [A,B]$; hence two maps $X\times A\to X\times B$ over $X$ are equal as soon as they become so after pulling back along $\flat X\to X$. Right? Why does that imply that a single map $X\times A\to X\times B$ over $X$ is an equivalence as soon as it becomes so after pulling back to $\flat X$? *A priori* it seems that its inverse $\flat X\times B\to \flat X\times A$ might not come from any map over $X$.

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

7. • CommentRowNumber7.
   • CommentAuthorUrs
   • CommentTimeJan 15th 2015
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   Author: Urs
   Format: MarkdownItexI see, thanks!

   I see, thanks!