Clear all

# Adjunction of frames and its induced adjunction of presheaf topoi

(@matteocapucci)
Active Member
Joined: 5 months ago
Posts: 15
Topic starter

Suppose $f: \mathcal F \to \mathcal G$ is a morphism of frames which has both posetal adjoints, $f_! \dashv f \dashv f_*$. By posetal I mean they don't necessarily preserve meets/joins (while $f$ does, of course)

In this situation, $f$ induces an adjunction of geometric morphisms $\mathfrak h: \mathrm{Psh} \mathcal G \leftrightarrows \mathrm{Psh}\mathcal{F} : \mathfrak g$, as explained here. Normally we would get only an essential geometric morphism $\mathfrak g : \mathrm{Psh} \mathcal F \to \mathrm{Psh}\mathcal{G}$, but left exactness of $f$ allows us to conclude there's a geometric morphism $\mathfrak h$ left adjoint to this one.

All in all, this amounts to three functors:

1. $\mathbf f = - \circ f : \mathrm{Psh} \mathcal G \to \mathrm{Psh}\mathcal{F}$, which plays the role of inverse part in $\mathfrak g$ and direct part in $\mathfrak h$
2. $\mathbf f_! = \mathrm{Lan}_f :\mathrm{Psh} \mathcal F \to \mathrm{Psh}\mathcal{G}$ which is left adjoint to of $\mathbf f$ and plays the role of inverse part in $\mathfrak h$
3. $\mathbf f_* = \mathrm{Ran}_f:\mathrm{Psh} \mathcal F \to \mathrm{Psh}\mathcal{G}$ which is right adjoint to of $\mathbf f$ and plays the role of direct part in $\mathfrak g$

First question: is the above account correct?

Then then question is, how are the two Kan extensions computed? In particular, since I'm dealing with frames, I should be able to get an explicit formula.

So my reasoning went like this: pretend $\mathcal F$ and $\mathcal G$ are frames of open subsets of a locale (well, there's nothing to pretend, they are), then we know an explicit formula, at least for the inverse part of $\mathfrak h$ (usually called the 'inverse image'):

$\mathbf f_!(F)(a) = \mathrm{colim}_{\mathcal F \ni b \leq a} F(b)$

Dually, I think we should get

$\mathbf f_*(F)(a) = \mathrm{lim}_{\mathcal F \ni b \geq a} F(b)$

Is this correct?

The final question concerns the relationship between $\mathbf f_!$ and $\mathbf f_*$ and the posetal adjoints to $f$. At first I though that maybe

$\mathbf f_!(F)(a) = F(f f_! (a))$

$\mathbf f_*(F)(a) = F(f f_*(a)),$

But now I'm not very convinced since it would imply some sort of commutation of $F$ with co/limits, which doesn't seem really plausible in general. Maybe it works when $F$ is a sheaf? (with respect to jointly surjective coverings).

This topic was modified 5 months ago by MatteoCapucci

Topic Tags
(@edenstar)
New Member
Joined: 5 months ago
Posts: 4

@matteocapucci I believe that your account of the adjoint triple on the presheaf categories is correct. To get an explicit (co)limit formula for them you can use the (co)end formulas for left and right Kan extensions. You can find these formulas here: https://ncatlab.org/nlab/show/Kan+extension#PointwiseByCoEnds

(@edenstar)
New Member
Joined: 5 months ago
Posts: 4

Note that those formula are for arbitrary enrichment, but because we're talking about Set enrichment, tensor is cartesian product and the hom-object is the usual hom-set.

This post was modified 5 months ago by edenstar

(@edenstar)
New Member
Joined: 5 months ago
Posts: 4

Also, I am pretty sure that you can get the co(limit) formulas which apply to posets of open sets of a space as a special case. If your categories are posets then the cartesian product in the coend formula is either with the one-element set or the empty set. So you end up taking the colimit over all objects which have a morphism into a i.e. over all subsets of a.

(@jwrigley)
Active Member
Joined: 5 months ago
Posts: 13

Hi Matteo, your account on the geometric morphisms is correct (since the essential part $f_!$ is left exact).

Then I'm a little confused by your colimit notation.

Posted by: @matteocapucci

we know an explicit formula, at least for the inverse part of $\mathfrak h$ (usually called the 'inverse image'):

$\mathbf f_!(F)(a) = \mathrm{colim}_{\mathcal F \ni b \leq a} F(b)$

If $F$ is a functor $F \colon \mathcal{F}^{op} \to {\bf Sets}$ then shouldn't $a \in \mathcal{G}$? So how can you take the colimit over $\mathcal{F} \ni b \leqslant a$?

Since $f_!$ is the left Kan extension of ${\bf y}_\mathcal{G} \circ f$ (SGL, pg. 417) we have that $f_! \circ {\bf y}_{\mathcal{F}} = {\bf y}_{\mathcal{G}} \circ f$.  Moreover $F$ is a colimit of representables (SGL, pg. 63) and, being a left adjoint, $f_!$ preserves that colimit.  Hence we have that $f_!(F)$ is the colimit of the composite $\int F \xrightarrow{\pi_F} \mathcal{F} \xrightarrow{{\bf y}_\mathcal{F}} {\bf Sets}^{\mathcal{F}^{op}} \xrightarrow{- \circ f} {\bf Sets}^{\mathcal{G}^{op}}$, which can be computed pointwise in the topos of presheaves.  Does that help?

(@matteocapucci)
Active Member
Joined: 5 months ago
Posts: 15
Topic starter
Posted by: @edenstar

@matteocapucci I believe that your account of the adjoint triple on the presheaf categories is correct. To get an explicit (co)limit formula for them you can use the (co)end formulas for left and right Kan extensions. You can find these formulas here: https://ncatlab.org/nlab/show/Kan+extension#PointwiseByCoEnds

Good idea, I didn't think of using the co/end formulas. Checking them, I get what I ansatzed in my original post

(@matteocapucci)
Active Member
Joined: 5 months ago
Posts: 15
Topic starter
Posted by: @jwrigley

Hi Matteo, your account on the geometric morphisms is correct (since the essential part $f_!$ is left exact).

Then I'm a little confused by your colimit notation.

Posted by: @matteocapucci

we know an explicit formula, at least for the inverse part of $\mathfrak h$ (usually called the 'inverse image'):

$\mathbf f_!(F)(a) = \mathrm{colim}_{\mathcal F \ni b \leq a} F(b)$

If $F$ is a functor $F \colon \mathcal{F}^{op} \to {\bf Sets}$ then shouldn't $a \in \mathcal{G}$? So how can you take the colimit over $\mathcal{F} \ni b \leqslant a$?

Good catch! It's a typo (sigh, also ! and * should be superscripts, not subscripts), the formula should read:
$\mathbf f_!(F)(a) = \mathrm{colim}_{\mathcal F \ni f(b) \leq a} F(b)$
Same for the right Kan extension.

Also I wanted to add that the co/end formulation of Kan extensions suggested by @edenstar also answers the second question of mine, namely how to give an explicit formula for $\mathbf f_!(F)(a)$ in terms of the adjoints of $f$. If you apply said adjunction inside the co/end formulas and then use the appropriate Yoneda reduction you get to

$\mathbf f_!(F)(a) = F(f_*(a))$

$\mathbf f_*(F)(a) = F(f_!(a))$

This post was modified 5 months ago by MatteoCapucci

(@john-dee)
Active Member
Joined: 5 months ago
Posts: 5

@matteocapucci You already found a way to rewrite the left and right Kan extensions of $f$; if you're interested in the bigger picture, the request that $f$ has posetal adjoints on both sides implies that $f$ represents a particular open map of locales; Elephant C.1.5.3 now gives you equivalent conditions for openness in terms of the induced geometric morphism.

(@matteocapucci)
Active Member
Joined: 5 months ago
Posts: 15
Topic starter

Thanks @john-dee, that should be relevant to another question of mine that I didn't communicate here.

Share: