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Suppose that $\mathcal{T}$ is an abstract $2$-category we know is equivalent to the $2$-category of Grothendieck topoi via some equivalence $$\phi:\mathcal{T} \to \mathfrak{Top},$$ and let $E$ be an object of $T$. Can we recover the underlying category $\phi(E)$ without using $\phi$?

I am asking because often properties of morphisms of topoi use that we know what topoi are (certain categories) and what maps between them are (certain pairs of adjoint functors), e.g. by referencing elements of the domain, or by saying one of the pairs of adjoint functors has a further adjoint with certain properties. But, a general principle of category theory is that one shouldn't care what things are, just about the maps between them-and not really the maps, but how they are related, i.e. the category you get. If you have an equivalent category, then you should be able to make the same statements. So the $2$-category $\mathcal{T}$ should be enough. Hence, I ask, how does one recover $\phi(E)$? It would suffice have a completely categorical description of etale geometric morphisms (i.e. one not depending on the "evil" choice of a particular presentation of $\mathfrak{Top}$ as consisting of certain categories called topoi) because any topos is equivalent to the category of etale morphisms over itself.

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    $\begingroup$ I have always found the "evil" terminology to be unfortunate. This regrettable terminology seems to hinder and sometimes to halt discussion between category theorists and those outside category theory by burdening the conversation with unnecessarily confrontational morally-loaded language. Why not adopt a more neutral and descriptive term? $\endgroup$
    – Joel David Hamkins
    Commented Apr 4, 2013 at 3:02
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    $\begingroup$ While I don't mind the "evil" terminology quite as much as JDH, it always reminds me of (a) Bill Bailey's "Scale of Evil" (b) how much I liked learning category theory from someone who avoided overtones of That Was Wrong Now This Is Right $\endgroup$
    – Yemon Choi
    Commented Apr 4, 2013 at 3:10
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    $\begingroup$ You have to understand that "evil" was originally just an in-joke, originating from UC Riverside I think. The regrettable thing is how it broke into public discourse, probably mostly via the internet, so that the tongue-in-cheekiness was no longer so apparent. There was furious backlash against it and endless discussion about it within categorical circles (especially from older generations of category theorists). More discussion here: ncatlab.org/nlab/show/principle+of+equivalence $\endgroup$
    – Todd Trimble
    Commented Apr 4, 2013 at 11:47
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    $\begingroup$ David, I don't think that reasoning works, unless you also believe that category theory has no applications outside itself. $\endgroup$
    – François G. Dorais
    Commented Apr 4, 2013 at 15:21
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    $\begingroup$ Re JDH's question: the more neutral term adopted by the nLab is "principle of equivalence". It is indeed a useful general theoretical principle, but there are significant cases where the rule is sensibly broken (I mean broken even on the theoretical level, not just the practical or calculational level where the rule is frequently broken as a convenience). Understanding these issues is tricky enough; I agree a loaded term like "evil" doesn't help anyone's understanding. (It was a joke that got out of hand.) $\endgroup$
    – Todd Trimble
    Commented Apr 4, 2013 at 15:49

1 Answer 1

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There is a Grothendieck topos $\textbf{Set}[\mathbb{O}]$ with the following universal property: for all Grothendieck toposes $\mathcal{E}$, the category $\textbf{Geom}(\mathcal{E}, \textbf{Set}[\mathbb{O}])$ of geometric morphisms $\mathcal{E} \to \textbf{Set}[\mathbb{O}]$ and "geometric transformations" (a misnomer; they actually code algebraic data!) is naturally equivalent to $\mathcal{E}$ itself. Such a $\textbf{Set}[\mathbb{O}]$ is called an object classifier.

We can construct $\textbf{Set}[\mathbb{O}]$ explicitly using the theory of classifying toposes: one presentation is as the presheaf topos $[\textbf{FinSet}, \textbf{Set}]$. Indeed, by Diaconescu's theorem, a geometric morphism $\mathcal{E} \to [\textbf{FinSet}, \textbf{Set}]$ is the same thing as a left exact functor $\textbf{FinSet}^\textrm{op} \to \mathcal{E}$, but any such is freely and uniquely determined by the image of $1$; this correspondence extends to 2-morphisms as well.

Addendum. To address Simon Henry's comments, here is an abstract construction of $\textbf{Set}[\mathbb{O}]$. It is known that the 2-category of Grothendieck toposes has tensors with small categories. Indeed, $$[\mathbb{C}, \textbf{Geom}(\mathcal{E}, \mathcal{F})] \simeq \textbf{Geom}([\mathbb{C}, \mathcal{E}], \mathcal{F})$$ but we know that $\textbf{Set}$ is a pseudo-terminal object in the 2-category of Grothendieck toposes, so we may take $\textbf{Set}[\mathbb{O}] = \textbf{FinSet} \otimes \mathcal{S}$, where $\mathcal{S}$ is any pseudo-terminal Grothendieck topos.

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  • $\begingroup$ Yes but you need to know $\phi(E)$ in order to compute $Geom(\phi(E),Set[\mathbb{O}])$, no ? $\endgroup$
    – Simon Henry
    Commented Apr 4, 2013 at 7:45
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    $\begingroup$ I mean't you need to assume that you know which object of $\mathcal{T}$ is equivalent to $Set[\mathbb{O}]$. I don't think there is an actual 'universal property' defining him... $\endgroup$
    – Simon Henry
    Commented Apr 4, 2013 at 8:01
  • $\begingroup$ Yes, but how is this any different from needing to know which object is $\mathbb{Z}$ in $\textbf{Ab}$ in order to recover the underlying set of an abelian group? $\endgroup$
    – Zhen Lin
    Commented Apr 4, 2013 at 8:09
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    $\begingroup$ In Ab, the object $\mathbb{Z}$ can be characterized in purely categorical term : for example it is the only projective object $Z$ such that any other projective can be written as a coproduct of a familly of copies of $Z$. I don't know if $Set[\mathbb{O}]$ can be characterized similarly in a purely categorical way. (or equivalently, does every self-equivalence of the 2-category of toposes preserve $Set[\mathbb{O}]$ up to equivalence). This question is actually equivalent to the the initial questions. $\endgroup$
    – Simon Henry
    Commented Apr 4, 2013 at 9:48
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    $\begingroup$ I.e. the functor $$\mathcal{T} \to \mathbf{Cat}$$ sending $E$ to $$Nat\left(\mathbf{FinSet}^{op},\mathcal{T}\left(1,E\right)\right)$$ is corepresentable by an object $\mathcal{O}$ such that for each $E$ in $\mathcal{T}$, $\mathcal{T}\left(\mathcal{O},E\right)$ is equivalent to $\phi\left(E\right)$. Nice! $\endgroup$
    – David Carchedi
    Commented Apr 4, 2013 at 15:28

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