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Stream: deprecated: topos theory

Topic: left adjoint to sheafification

Matteo Capucci (he/him) (Nov 24 2020 at 20:21):

Is there some general useful fact about the existence of left adjoints to sheafification? This is equivalent to ask what are the levels of a presheaf topos
Anyway, I think we can go straight to the point here: I have a complete Boolean algebra, and I'd like to show sheaves over it (wrt the usual coverage of jointly surjective families) are a level of the presheaves over it. I need that left adjoint to sheafification!

sarahzrf (Nov 25 2020 at 06:52):

full subcategories induce levels of a presheaf topos—idk whether those are all of them

sarahzrf (Nov 25 2020 at 06:59):

i think this should be true for complete atomic boolean algebras at least

sarahzrf (Nov 25 2020 at 06:59):

i.e., when you take sheaves on a discrete space

sarahzrf (Nov 25 2020 at 07:00):

because then the category of sheaves is equivalently a power of Set, and sheafification can be seen as just forgetting the values of the presheaf on everything except singletons—and that definitely preserves limits, so between groth topoi, it's a right adjoint

sarahzrf (Nov 25 2020 at 07:03):

(...i need more practice thinking about sheafification for sites other than topological spaces...)

Matteo Capucci (he/him) (Nov 25 2020 at 10:53):

I actually realized that I need a right adjoint to the direct image, not a left adjoint to the inverse. A local geometric morphism would be enough but also too much, since locality asks quite a bit more and I'm not sure I can get it

Morgan Rogers (he/him) (Nov 25 2020 at 11:29):

Yes, you don't want something local! Rather, you're looking for a totally connected geometric morphism in the opposite direction? (A totally connected geometric morphism is a morphism which is connected (inverse image full and faithful), and whose extra left adjoint preserves finite limits)

Matteo Capucci (he/him) (Jan 06 2021 at 18:00):

Well, here I am again: I really need a left adjoint :laughing: @Jens Hemelaer and @Olivia Caramello confirmed your reasoning, @sarahzrf: I have it when the algebra is atomic

Matteo Capucci (he/him) (Jan 06 2021 at 18:01):

Now I'm struggling to understand how this left adjoint actually looks like so that I can compute it on some sheaves

Matteo Capucci (he/him) (Jan 06 2021 at 18:02):

It is here that I understand how noob I still am :sweat_smile: I tried helplessly for an hour to write down its expression using various general constructions of left adjoints but they all seem far removed from my situation and I can't really go past some large limits.

Matteo Capucci (he/him) (Jan 06 2021 at 18:08):

It'd be great to have some kind of arguments involving maps between $B$ (the atomic Boolean algebra I'm taking sheaves on) and its downset completion $B^\wedge$.

Matteo Capucci (he/him) (Jan 06 2021 at 18:29):

Maybe I can reason like this: since $B$ and $B^\wedge$ are frames, then I can use the fact that essential subtopoi correspond to essential sublocales, which in turn are given by left adjoints of nuclei $B^\wedge \to B^\wedge$. Hence since the nucleus on $B^\wedge$ inducing the embedding $Sh B \to Psh B$ is given by $j:D \mapsto \downarrow \bigvee D$ for $D$ downset, we are looking for a left adjoint to this.

Matteo Capucci (he/him) (Jan 06 2021 at 18:32):

Call this left adjoint $b$ as in the linked nLab page. Now I'm left with two problems: (1) find an expression for $b$, but this should be easy and it probably involves decomposing in atoms, and (2) using $b$ to write down a left adjoint to sheafification.

Jens Hemelaer (Jan 07 2021 at 10:06):

sarahzrf said:

because then the category of sheaves is equivalently a power of Set, and sheafification can be seen as just forgetting the values of the presheaf on everything except singletons—and that definitely preserves limits, so between groth topoi, it's a right adjoint

I think this is the key to explicitly write down the left adjoint to sheafification.

Suppose that the complete atomic boolean algebra $B$ is given as $B = \mathcal{P}(S)$. Note that $\mathbf{PSh}(S) \simeq \mathbf{Sh}(B)$. Now consider the presheaf on $S$ which is a singleton over some element $a \in S$ and the empty set elsewhere. Then the corresponding sheaf on $B$ is the representable presheaf $\mathbf{y}\{a\}$ where $\mathbf{y}$ denotes the Yoneda embedding.

I'll write $i^*$ for the sheafification and $i_!$ for its left adjoint. To compute $i_!$ it is enough to compute it on the sheaves of the form $\mathbf{y}\{a\}$, because all others are colimits (even coproducts) of these. Now:
$\mathrm{Hom}(i_!(\mathbf{y}\{a\}), \mathcal{F}) \simeq \mathrm{Hom}(\mathbf{y}\{a\}, i^*\mathcal{F}) \simeq (i^*\mathcal{F})(\{a\})$
Further, as @sarahzrf said, we have $(i^*\mathcal{F})(\{a\}) = \mathcal{F}(\{a\})$.
It follows that $i_!(\mathbf{y}\{a\}) = \mathbf{y}\{a\}$.

Matteo Capucci (he/him) (Jan 07 2021 at 17:48):

God, it really was quite easy :laughing:

Matteo Capucci (he/him) (Jan 07 2021 at 17:59):

Does this mean $i_!$ is again the inclusion $\mathbf{Sh}B \to \mathbf{Psh}B$? Since as a left adjoint it preserves colimits hence if it is the inclusion on the representables it's the inclusion on everything

Jens Hemelaer (Jan 07 2021 at 18:20):

Matteo Capucci said:

God, it really was quite easy :laughing:

At least in hindsight :smiley:

And no, I don't think $i_!$ is again the inclusion. For example:
$i_!(i^*(\mathbf{y}\{a,b\})) = \mathbf{y}\{a\} \sqcup \mathbf{y}\{b\}$.
To see that $\mathbf{y}\{a,b\} \neq \mathbf{y}\{a\} \sqcup \mathbf{y}\{b\}$, you can check the sections over $\{a,b\}$.

Matteo Capucci (he/him) (Jan 07 2021 at 18:23):

Oh god, of course. I really shouldn't do any topos theory this late in the day :face_palm:

Matteo Capucci (he/him) (Jan 07 2021 at 22:31):

Contravening to my previous conclusion about thinking about topos theory late in the day, I was thinking about this and something is off.
I fail to see why your remark that every sheaf on $B$ is a coproduct of $y\{a\}$. Every representable presheaf on $B$ should be a sheaf for the canonical topology, since $B$ is distributive. Then you have sheaves like $y\{a, b\}$ which, as you note, are not the sum of $y\{a\}$ and $y\{b\}$.
Since

$$i_*(F)(A) = \prod_{a \in A} F\{a\}$$

and hom is continuous on the left, following your same proof one gets

$$i_! (yA) = \sum_{a \in A} y\{a\}$$

which seems sensible.

Matteo Capucci (he/him) (Jan 07 2021 at 23:03):

For an arbitrary sheaf $F$ then one gets

Matteo Capucci (he/him) (Jan 07 2021 at 23:03):

$i_! (F) (A) = \sum_{a \in A} F\{a\}$

Jens Hemelaer (Jan 08 2021 at 09:53):

I think the problem here is that the coproduct of two sheaves is not computed by taking the coproduct of underlying presheaves.

Take the natural inclusion of presheaves $j : \mathbf{y}\{a\} \sqcup \mathbf{y}\{b\} \to \mathbf{y}\{a,b\}$. This is not a bijection, because the domain has no sections over $\{a,b\}$ and the codomain does. But after sheafification, this map becomes an isomorphism of sheaves. To see this, it is enough to show that it becomes a bijection at each stalk (because the topos has enough points). Each stalk is given by taking the sections over $\{c\}$ for some element $c \in S$. The map $j$ is given as follows on each stalk:

• If $c \notin \{a,b\}$, then both stalks are empty.
• If $c = a$ or $c = b$, then both stalks are a singleton and $j$ restricts to a bijection.

In the same way, you can show that every sheaf on $B$ is a coproduct of sheaves of the form $\mathbf{y}\{x\}$ (in the category of sheaves, not in the category of presheaves). But there is an easier proof using $\mathbf{Sh}(B) \simeq \mathbf{PSh}(S)$. First prove that every presheaf in $\mathbf{PSh}(S)$ is a coproduct of representable presheaves. Then show that the representable presheaves in $\mathbf{PSh}(S)$ correspond to the sheaves of the form $\mathbf{y}\{x\}$ in $\mathbf{Sh}(B)$.