The n-Category Café

I hadn’t yet attempted to fix that bunch of mistakes, just the ones Todd and Tim pointed out. [Note added later: I have now tried to fix those mistakes.]

I am constantly confused about what people call $f^\ast$ and what people call $f_\ast$ in topos theory, in part because the ‘geometric’ way of saying which way geometric morphisms point is the opposite of the (to me more natural) ‘algebraic’ way.

If I have a functor $f: C \to D$, to me the most obvious map between presheaf categories is precomposition with $f$, which gives a functor $\widehat{D} \to \widehat{C}$. I would naturally call this $f^\ast$ because of the contravariance. But precisely because that’s what I want to do, I immediately suspect I should do the opposite!

(This is why I almost got run over in England: my mnemonic for traffic directions was “the cars are coming the opposite way from what you expect”, which worked until my expectations changed.)

And when I look at the nLab, my fears seem to be confirmed. It seems to say that this “precomposition with $f$” functor is called $f_\ast$:

Given a morphisms of sites $f : X \to Y$ coming from a functor $f^t : S_Y \to S_X$ of the underlying categories, the direct image operation $f_* : PSh(X) \to PSh(Y)$ on presheaves is just precomposition with $f^t$.

Admittedly, they are talking about precomposition with something called $f^t$, not $f$. They seem to have stuck in an algebraic/geometric reversal here by having arrows in some category of sites point the opposite direction from functors between the underlying categories. So I thought their $f^t$ is what I’d call $f$, while their $f$ is some clever formal trick. (Is it really helpful to introduce both $f$ and $f^t$ here?)

As I read this passage in the nLab, I could see my chances of getting the conventions right dwindling before my very eyes. But it does seem to say that precomposition with a functor gives a functor between presheaf categories called “direct image”, not “inverse image”. But you’re saying it’s called inverse image? What’s going on?

The notion of “morphism of sites” is a mess, with multiple conventions and definitions. I suggest ignoring it while you are looking at presheaves, and stick to functors.

The main point (and emphasised by Olivia Caramello and Riccardo Zanfa recently), is that a functor induces an essential geometric morphism between presheaf toposes, pointing in the same direction (where “essential” means it has an extra left adjoint).

(This is why I almost got run over in England: my mnemonic for traffic directions was “the cars are coming the opposite way from what you expect”, which worked until my expectations changed.)

Yes, I know the phenomenon you mean with “it’s the opposite of what you think” mnemonics.

I also have a problem with traffic in England. When I go there, I tend to have the sensation that I’ve travelled abroad. This means that as I step off the kerb, I instinctively look the opposite way from the way I normally would. Disaster.

In phases of my life when I’m travelling a lot, I try to drill myself to look both ways both at home and away. Apart from the England phenomenon, I think I’m the greatest risk to myself just after I’ve got home.

Not that the analogous move would help mathematically… Anyway, the convention is that for a functor $f: C \to D$, precomposition with $f$ is written as $f^\ast: \hat{D} \to \hat{C}$, its right adjoint (right Kan extension along $f$) is written as $f_\ast$, and its left adjoint is written as $f_!$. So $f_! \dashv f^\ast \dashv f_\ast$. The adjunction $f^\ast \dashv f_\ast$ is a geometric morphism $\hat{C} \to \hat{D}$, with $f^\ast$ being the inverse image part and $f_\ast$ the direct image part.

I have a “falling star” mnemonic, so reading left to right the star falls to the ground: $f^\ast \dashv f_\ast$.

Yes, the geometric morphism $F: \widehat{C} \to \widehat{C+D}$ is atomic (or in other words, its inverse image functor, given by restriction, is logical).

Here is a “geometric” kind of proof.

We can first write $\widehat{C + D} = \widehat{C} \sqcup \widehat{D}$ where $\sqcup$ denotes the coproduct in the category of toposes. Then we can write $\widehat{C} \to \widehat{C} \sqcup \widehat{D}$ as the pullback of $\widehat{1} \to \widehat{1} \sqcup \widehat{1}$ along $\widehat{C} \sqcup \widehat{D} \to \widehat{1} \sqcup \widehat{1}$.

Because atomic morphisms are stable under pullback, it is enough to show that $\widehat{1} \to \widehat{1} \sqcup \widehat{1}$ is atomic. But this is the geometric morphism induced by the inclusion of a point in the discrete two-point space, which is a local homeomorphism, in particular atomic.

Alternatively, you can show that $\widehat{C}$ can be written as the slice topos $\widehat{C+D}/Y$ for some object $Y$ in $\widehat{C+D}$. (Take $Y(c)=1$ for $c \in C$ and $Y(d)=\varnothing$ for $d \in D$.) The projection from the slice topos to the base topos is a local homeomorphism, so in particular atomic.

Isn’t the inverse image in this case just a projection? I.e., we have

$$\widehat{C + D} \simeq \widehat{C} \times \widehat{D}$$

and the restriction along the inclusion $C \to C + D$ is thereby just the projection

$$\widehat{C} \times \widehat{D} \to \widehat{C}.$$

But it’s pretty automatic that the projection will be logical, since limits, exponentials, and the subobject classifier are computed componentwise.

I just noticed that the direct image functor $F_*$ is sublogical if and only if the geometric morphism $F$ is an inclusion… (C3.1.8 in the Elephant, as Sam Staton pointed out). This agrees with what Peter Scholze mentioned for topological spaces.

So my examples above are a bit far-fetched. Any fully faithful functor $f: C \to D$ between small categories induces an inclusion $F : \widehat{C} \to \widehat{D}$, and these then all have sublogical direct image functor. On the other hand, if the direct image functor is logical, then in particular it preserves the subobject classifier, so $F$ is hyperconnected (A.4.6.6 in the Elephant). For inclusions this is only possible if $F$ is an equivalence (because hyperconnected geometric morphisms are surjective).

So any fully faithful functor $f : C \to D$ should work, as long as the induced geometric morphism $F : \widehat{C} \to \widehat{D}$ is not an equivalence.