Let $\mathcal{E}$ and $\mathcal{F}$ be toposes, $X$ an object of $\mathcal{E}$ and $p: \mathcal{E}/X \rightarrow \mathcal{E}$ the canonical geometric morphism (whose inverse image part is pullback along $p: X \rightarrow 1$.) I am trying to figure out the correspondence between geometric morphisms $\mathcal{F} \rightarrow \mathcal{E}/X$ over $\mathcal{E}$ and "global sections" $1 \rightarrow f^*(X)$. More precisely, I am looking to establish $$\text{Hom}_{\textbf{Topoi}/ \mathcal{E}} ( f: \mathcal{F} \rightarrow \mathcal{E}, \mathcal{E}/X) \cong \text{Hom}_{\mathcal{F}} (1, f^{*} (X))$$

Given a geometric morphism $g$ it is easy to get a section. One simply sends $g$ to the section $$g^* (\Delta) : 1 \rightarrow g^*(\pi_2 : X \times X \rightarrow X) = g^* \circ p^* (X) = f^*(X) $$

where $\pi_2$ is the standard projection and $\Delta$ the diagonal map $X \rightarrow X \times X$. (The idea is that $\Delta$ is the universal section, classifying the identity $\mathcal{E}/X = \mathcal{E}/X$.)

I'm having some difficulty figuring out the converse. Given a section $s : 1 \rightarrow f^*(X)$ we do get a geometric morphism $\mathcal{F} \rightarrow \mathcal{F}/f^*(X)$ and I can see that this ought to be 'sendable' to $\mathcal{E}/X$ but cannot figure out a nice way to do this via a geometric morphism (the wall I keep hitting is that the unit map $\eta_X : X \rightarrow f_* f^*(X)$ induces a geometric morphism in the 'wrong' direction.)

So I guess my first question is:

How does one get the right-to-left correspondence, i.e. from sections to geometric morphisms in this case?

In general, I am also interested to hear how one should generally think of these classifying problems (finding classifying topoi for given structures.) I am new to the game and every case I encounter seems to me now devoid of a general principle of attack. In some cases it seems much easier to go from 'points'/'individual structures' to geometric morphisms. For example, in geometric cases, a point immediately gives the stalk/skyscraper adjunction and hence a geometric morphism. In other cases, like the one above, where the universal structure is clear, it is much easier to go from geometric morphism to a 'point'/'structure'.

So my second question is this:

Is there a good standard way to think/attack such problems? Should one always try and figure out the universal structure first? Should one think geometrically? Any suggestions will be greatly appreciated...

(NOTE: I understand that the second question is not very precise, but neither am I looking for very precise answers.)

Finally, a bonus question: is the correspondence in the problem above (the first equation) a kind of adjunction between functors $\textbf{Topoi}/\mathcal{E} \overset{\leftarrow}{\rightarrow} \mathcal{F}$ (where $\mathcal{F}$ lies over $\mathcal{E}$.) Can one view this as some sort of adjunction in this manner? I suspect not, but I'd like to know if this is a fact made more elegant from a 2-categorical perspective...

  • $\begingroup$ Can you give a source for your claims? I've never seen anything like it. If we take the case $\mathcal{E} = \textbf{Set}$, $\mathcal{F} = \textbf{Set}$, then it's certainly true that an $\mathcal{E}$-morphism $\mathcal{F} \to \mathcal{E} / X$ corresponds to an element of $X$. But it feels like something much more complicated should happen in general... $\endgroup$ – Zhen Lin Apr 30 '12 at 8:07
  • $\begingroup$ @Zhen Lin This is exercise 6 of chapter VIII of Moerdijk and Maclane. (I didnt put it in the main text so as not to make the exercise searchable) $\endgroup$ – Chuck Apr 30 '12 at 12:29
  • $\begingroup$ That seems like a silly thing to do. Anyone who wants to search is going to search for keywords rather than an explicit exercise number! $\endgroup$ – Zhen Lin Apr 30 '12 at 15:56
  • $\begingroup$ @Zhen Yes, you're probably right... $\endgroup$ – Chuck Apr 30 '12 at 16:35

Allow me to change notation a little. Let $\mathcal{S}$ be a topos, and let $f : \mathcal{E} \to \mathcal{S}$ be a geometric morphism. Let $\mathcal{S}_X = (\mathcal{S} \downarrow X)$. We are trying to look at the hom-category $$\textbf{Geom}_{\mathcal{S}}(\mathcal{E}, \mathcal{S}_X)$$ which is the full subcategory of $\textbf{Geom}(\mathcal{E}, \mathcal{S}_X)$ spanned by geometric morphisms $g : \mathcal{E} \to \mathcal{S}_X$ such that $p \circ g \cong f$, where $p : \mathcal{S}_X \to \mathcal{S}$ is the canonical projection.

The situation we should keep in mind is the case where $\mathcal{E} = \mathcal{S}_Y$ for some other object $Y$. Then, we are just looking at base change along some morphism $Y \to X$ in $\mathcal{S}$. Geometric morphisms like $p : \mathcal{S}_X \to X$ are very special: they come as a triple of adjoint functors $$p_! \dashv p^* \dashv p_* : \mathcal{S}_X \to \mathcal{S}$$ and $p_! 1$ is canonically isomorphic to $X$. Unfortunately, $f : \mathcal{E} \to \mathcal{S}$ is not guaranteed to be of this form, so we must find a surrogate for $f_! 1$.

But, for the moment, let us imagine we had $f_! \dashv f^* \dashv f_*$ of the same form as $p_! \dashv p^* \dashv p_*$. Let $x : 1 \to f^* X$ be a morphism in $\mathcal{E}$. It would correspond to a unique morphism $f_! 1 \to X$ in $\mathcal{S}$. Take $Y$ in $\mathcal{S}_X$. It corresponds to a unique morphism $p_! Y \to X$ in $\mathcal{E}$. We want to find a left exact left adjoint functor $g^* : \mathcal{S}_X \to \mathcal{E}$, so there's only one thing we can do: take the base change of $p_! Y \to X$ along $f_! 1 \to X$. This "is" the required object in $\mathcal{E}$, except it's in $\mathcal{S}$ rather than in $\mathcal{E}$. Now, apply $f^*$ to the pullback diagram; since $f^*$ is left exact, we get a pullback diagram in $\mathcal{E}$. Consider the naturality square of the unit $\textrm{id} \Rightarrow f^* f_!$: if $f_! \dashv f^* \dashv f_*$ comes from base change, then this would be a pullback square. So by the pullback pasting lemma, we can get the base change of $p_! Y \to X$ along $f_! \to X$ as an object in $\mathcal{E}$ by taking the base change of $f^* p_! Y \to f^* X$ along $x : 1 \to f^* X$. Notice that we've erased all mention of $f_!$ by doing this!

But enough handwaving. Let's actually define the functor $g^*$ properly now. Write $x^* : \mathcal{E}_{f^* X} \to \mathcal{E}$ for base change along $x : 1 \to f^* X$. One may check that $f^* p_! : \mathcal{S}_X \to \mathcal{E}$ factors through the forgetful functor $\mathcal{E}_{f^* X} \to \mathcal{E}$; abusing notation slightly, let us also write $f^* p_!$ for the functor $\mathcal{S}_X \to \mathcal{E}_{f^* X}$. This is obviously left exact, so the composite $g^* = x^* f^* p_! : \mathcal{S}_X \to \mathcal{E}$ is also left exact. In less sophisticated terms, $g^*$ maps an object $Y$ of $\mathcal{S}_X$ (i.e. a morphism $p_! Y \to X$ of $\mathcal{S}$) to the object $1 \times_{f^* X} f^* p_! Y$ of $\mathcal{E}$ obtained by the base change of $f^* p_! Y \to f^* X$ along $x : 1 \to f^* X$. One may verify that $g^*$ preserves all colimits, so it has a right adjoint by the special adjoint functor theorem. It is easy to see that $g^* \cong f^* p^*$, so we have the desired geometric morphism.

Your second question about how to find classifying toposes deserves a good answer, but I'm afraid I don't have the time to write one right now. Since it's not quite related to the first question, perhaps it would be best to start a new thread on it and see if anyone else answers.

  • $\begingroup$ Thanks for this answer. I don't understand what version of the SAFT you are using there (which I associate with the existence of a left adjoint)? I'm assuming what you have in mind is the theorem that says that a functor between Grothendieck toposes has a right adjoint iff it preserves colimits. However we cannot know that the toposes in question here are Grothendieck. Maybe I'm missing something.... $\endgroup$ – Chuck May 4 '12 at 17:26
  • $\begingroup$ Yes, that's basically a dual form of the special adjoint functor theorem. You're right that we need to assume enough colimits + a "small" generating set, but I imagine an $\mathcal{S}$-indexed version of SAFT should be enough. Or you could construct the right adjoint by hand. shrug $\endgroup$ – Zhen Lin May 4 '12 at 17:57

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