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Does the 2-category of Grothendieck topoi have exponential objects?

There are size issues: Since Grothendieck topoi are supposed to have a small set of generators, the collection of objects of a Grothendieck topos has to be a class and cannot be a conglomerate. However, classes (in contrast to conglomerates) don't have nice closure properties: if $A$ and $B$ are classes, then the conglomerate $B^A$ is, in general, not a class.

So for the purposes of my question, drop the small set of generators from the Giraud axioms. :-)

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No. Some, but not all topoi are exponentiable.

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  • $\begingroup$ Thanks! When speaking of exponentiable toposes, does the nLab use the convention of dropping the small set of generators from Giraud's axioms as well? Otherwise I guess no large Grothendieck topos can be exponentiable. $\endgroup$
    – user1009651
    Commented Jan 3, 2022 at 16:51
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    $\begingroup$ @user1009651 No, the nlab is talking about Grothendieck toposes in the usual sense. The categorical product of Grothendieck toposes is not the same as the product of the underlying categories (in fact, the product of underlying categories is the coproduct in the 2-category of Grothendieck toposes and geometric morphisms), and the exponential $\mathcal F^{\mathcal E}$, when it exists, does not agree with the functor category or even the category of geometric morphisms from $\mathcal E$ to $\mathcal F$. $\endgroup$
    – Tim Campion
    Commented Jan 3, 2022 at 19:32
  • $\begingroup$ Whoops, right, what I really intended to ask was about possible right adjoints of $\mathcal E+-$, then. Do you have any idea concerning that? $\endgroup$
    – user1009651
    Commented Jan 4, 2022 at 16:40
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    $\begingroup$ The functor E+ fails to preserve the initial object (unless E is the initial topos itself) so it doesn’t have a right adjoint. $\endgroup$
    – Tim Campion
    Commented Jan 4, 2022 at 21:31
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    $\begingroup$ @user1009651 If $E$ and $F$ are Grothendieck toposes, then the conglomerate of geometric morphisms $E \to F$ is at least in bijection with a class (thought it may not technically be a class, I'm not sure). In fact, the larger conglomerate of all accessible functors $E \to F$ is in bijection with a class, because each accessible functor between accessible categories is determined up to isomorphism by what it does on a small subcategory, and the isomorphism classes of accessible functors are also class-sized. $\endgroup$
    – Tim Campion
    Commented Jan 5, 2022 at 18:13

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