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Stream: learning: questions

Topic: Cotensors in enriched presheaf categories

Ari Rosenfield (Mar 12 2025 at 01:43):

Say $\mathcal{V}$ is a symmetric monoidal closed category which is bicomplete and locally finitely presentable (so admitting a dense generating set of objects $\mathcal{G}$), and $\mathcal{C}$ is a $\mathcal{V}$-category. Suppose that $R,P \in [\mathcal{C}^\text{op},\mathcal{V}]$ are $\mathcal{V}$-functors.

For an arbitrary $g \in \mathcal{G}$, recall that the cotensor $\{g,P\}$ in $[\mathcal{C}^\text{op},\mathcal{V}]$ sends an object $x$ to $\mathcal{V}(g,Px)$. Given a $\mathcal{V}$-natural transformation $\alpha : R \to \{g,P\}$, is there a canonical way to extend $\alpha$ to a morphism $\{g,R\} \to \{g,P\}$?

I've been messing around with the relevant universal properties for a few hours and can't seem to make it happen. This is in the context of trying to generalize a lemma regarding sheaves with respect to some coverage (Lemma 3.2.10 from Borceux's Categorical Algebra III) to the enriched setting, using the definition of enriched coverage/Grothendieck topology found in this paper.

Morgan Rogers (he/him) (Mar 12 2025 at 13:04):

Is there a reason you expect this to be possible for general $g$? The simplest case is to take $\mathcal{V}$ to be $\mathrm{Set}$ and $\mathcal{C}$ to be the one-object category. Then cotensoring is exponentiation, and you're asking given $X\to Y^Z$ if there is a canonical extension to $X^Z \to Y^Z$. I don't think there is an extension you could define here that is natural in all three variables.

Ari Rosenfield (Mar 12 2025 at 13:52):

You're right, thanks! I had expected the proof in the unenriched case to work pretty much exactly as written for the enriched case, so I wasn't looking for counterexamples yet.

Morgan Rogers (he/him) (Mar 12 2025 at 15:23):

I still am curious if you have come across a case where this extension is possible, since it is an interesting possibility. :innocent:

Ari Rosenfield (Mar 12 2025 at 16:06):

No, I haven't. For more context, what I'm trying to do is this:

Given $P$, $g$, and $\alpha$ as above, a monomorphism $r : R \to \mathcal{C}(-,x)$, and also given a generalized element $f : g \to \mathcal{C}(y,x)$, I can take a pullback in $[\mathcal{C}^\text{op},\mathcal{V}]_0$, as in region (i) below. I was hoping to find a dashed arrow arising from $\alpha$, but I'm not very fluent with cotensors/generating families yet, so it's a struggle at the moment.

Untitled.png

Clearly, having the extension that I asked for in the original post would help, but you're right that it doesn't exist in general :joy:

Morgan Rogers (he/him) (Mar 12 2025 at 20:34):

I want to say I see, but I don't know if I'm supposed to see a relationship between the pullback square and the extension you're trying to find. On the other hand, if you instead had a morphism $R \to P$ the task would be very easy!
I would add that the existence of extensions (especially universal ones) defines a special property of either objects of [C^op,V] or of V (depending on which parameters are fixed/universally quantified).

Ari Rosenfield (Mar 12 2025 at 22:30):

I want to say I see, but I don't know if I'm supposed to see a relationship between the pullback square and the extension you're trying to find.

Yes, sorry, I think it probably looks pretty random if you haven't recently looked at the paper I'm referring to and/or the proof I'm trying to replicate. It all seemed like a lot for a Zulip post; I was trying to be concise.

On the other hand, if you instead had a morphism $R→P$ the task would be very easy!

In the unenriched case, that's exactly what happens!

I'll try to give a little more detail here in a second

Ari Rosenfield (Mar 12 2025 at 23:07):

By "enriched coverage," I mean a collection of sieves on objects of $\mathcal{C}$ which is closed under pullbacks (the paper I linked to in the first post has a definition of enriched topology, so in terms of that definition, it would be an enriched topology minus condition (T3)).

The definition of "enriched sheaf" I'm using is nearly the same as the unenriched version, but now I have to care about generalized elements: Say $P$ is a sheaf for a $\mathcal{V}$-coverage $J$ if there exists a unique $\beta$ such that for all $(r: R \rightarrowtail \mathcal{C}(-,x)) \in J(x)$, $g \in \mathcal{G}$, and $\alpha : R \to \{g,P\}$, this triangle commutes:

Untitled2.png

I'm trying to prove that any sheaf for $J$ is a sheaf for the (enriched) coverage defined on $x \in \mathcal{C}$ by $J^+(x) = \{(R \rightarrowtail \mathcal{C}(-,x)) : \exists \; S \in J(x) \text{ s.t. } \forall f \in \hom(\mathcal{G},S(y)), \; R_f \in J(y)\}$.

Given such an $S$, if I can find a dashed arrow as in the diagram from a couple messages back, $P$ being a sheaf for $J$ gives me a unique $\beta_f$ such that the following commutes:

Untitled.png

Then I think I can move on with the existence part of the proof. Even further out, this would be part of an inductive proof that sheaves for an enriched coverage $J$ are the same as those for the smallest enriched topology containing $J$. Obvi I need to do this first, though :sweat_smile:

Morgan Rogers (he/him) (Mar 13 2025 at 08:26):

Ari Rosenfield said:

The definition of "enriched sheaf" I'm using is nearly the same as the unenriched version, but now I have to care about generalized elements: Say $P$ is a sheaf for a $\mathcal{V}$-coverage $J$ if there exists a unique $\beta$ such that for all $(r: R \rightarrowtail \mathcal{C}(-,x)) \in J(x)$, $g \in \mathcal{G}$, and $\alpha : R \to \{g,P\}$, this triangle commutes:

Untitled2.png

The quantifiers seem to be the wrong way around in this sentence: $\beta$ should depend on $\alpha$, no?

Morgan Rogers (he/him) (Mar 13 2025 at 09:44):

I see, so you're trying to prove that the closure of the sieves under a kind of multicomposition yields the same sheaves. I would keed to think a lot more deeply to understand why the cotensor appears in the sheaf condition in the enriched setting (this surely shows my ignorance of the subtleties of enriched sheaves); could you explain that? Actually, it's not clear to me that the sheaf condition you give reduces to the usual one in the unenriched setting!

Ari Rosenfield (Mar 13 2025 at 17:10):

The quantifiers seem to be the wrong way around in this sentence: $β$ should depend on $α$, no?

oop yes, I've fixed it now!

Actually, it's not clear to me that the sheaf condition you give reduces to the usual one in the unenriched setting!

In the unenriched case, we can take $\mathcal{G}$ to be a single-element set. Then for all $g \in \mathcal{G}$ (i.e., for $g = \mathbf{1}$), we have $\{g, P\} \cong P$. The enriched sheaf condition is then equivalent to saying that for all sieves $S \in J(x)$, we have $\hom(S,P) \cong \hom(\mathcal{C}(-,x), P)$. If you buy that a matching family for $S$ of elements of $P$ is the same thing as a natural transformation $S \to P$, this is just saying that any matching family has a unique amalgamation (just following the argument on p.121-122 in Maclane-Moerdijk).

I would keed to think a lot more deeply to understand why the cotensor appears in the sheaf condition in the enriched setting (this surely shows my ignorance of the subtleties of enriched sheaves); could you explain that?

The enriched sheaf condition just takes the perspective that $\mathcal{V}$-transformations $S \to \{g,P\}$ are "matching families of generalized elements of shape $g$" of $P$. For a general $\mathcal{V}$, unlike the case of sets, "elements" $I \to P$ don't capture enough information about objects anymore, so we have to consider more general shapes.

Morgan Rogers (he/him) (Mar 13 2025 at 17:22):

Ari Rosenfield said:

Actually, it's not clear to me that the sheaf condition you give reduces to the usual one in the unenriched setting!

In the unenriched case, we can take $\mathcal{G}$ to be a single-element set. Then for all $g \in \mathcal{G}$ (i.e., for $g = \mathbf{1}$), we have $\{g, P\} \cong P$. The enriched sheaf condition is then equivalent to saying that for all sieves $S \in J(x)$, we have $\hom(S,P) \cong \hom(\mathcal{C}(-,x), P)$. If you buy that a matching family for $S$ of elements of $P$ is the same thing as a natural transformation $S \to P$, this is just saying that any matching family has a unique amalgamation (just following the argument on p.121-122 in Maclane-Moerdijk).

My point was rather that if $g$ is not a singleton, and more specifically if $g$ is not finite, then I don't think the extension property holds in the unenriched case, but here you're saying it should hold for arbitrary $g$!

Morgan Rogers (he/him) (Mar 13 2025 at 17:25):

Well I suppose you did say it had to belong to $\mathcal{G}$ but I could take that dense generating set to be arbitrarily large.

Morgan Rogers (he/him) (Mar 13 2025 at 17:27):

This would suggest to me that one should impose a compactness condition on the objects $g$, which if strong enough could allow you to construct the extension you're looking for (although the extension probably won't be unique/natural)

Ari Rosenfield (Mar 13 2025 at 17:53):

Yes, I think since $\mathcal{V}$ is locally finitely presentable, I can always take $\mathcal{G}$ to consist only of finitely presentable objects.

It's still not clear to me how to construct the extension, though.

Martti Karvonen (Mar 14 2025 at 08:56):

Morgan Rogers (he/him) said:

Is there a reason you expect this to be possible for general $g$? The simplest case is to take $\mathcal{V}$ to be $\mathrm{Set}$ and $\mathcal{C}$ to be the one-object category. Then cotensoring is exponentiation, and you're asking given $X\to Y^Z$ if there is a canonical extension to $X^Z \to Y^Z$. I don't think there is an extension you could define here that is natural in all three variables.

Given $f\colon X\to Y^Z$, can't we extend it to $\bar{f}\colon X^Z \to Y^Z$ by defining for $g\in X^Z$ the function $\bar{f}(g)\in Y^Z$ as $\bar{f}(g)(z):=f(g(z))(z)$? Seems like a canonical choice to me, although it might fail to be natural in all of the three variables as you say.

Morgan Rogers (he/him) (Mar 14 2025 at 09:57):

That's a sensible answer, I hadn't thought about diagonalizing. Nice one @Martti Karvonen

Morgan Rogers (he/him) (Mar 14 2025 at 09:59):

Morgan Rogers (he/him) said:

My point was rather that if $g$ is not a singleton, and more specifically if $g$ is not finite, then I don't think the extension property holds in the unenriched case, but here you're saying it should hold for arbitrary $g$!

In case it wasn't clear why I don't expect it to hold for arbitrary $g$, I think this would imply that sheafification preserves some infinite powers of presheaves, and this shouldn't be the case in general.

Ari Rosenfield (she/her) (Mar 14 2025 at 15:12):

Thank you so much @Morgan Rogers (he/him) and @Martti Karvonen for helping me think through this!

Given $f ⁣:X→YZ$, can't we extend it to $fˉ ⁣:XZ→YZ$ by defining for $g∈XZ$ the function $fˉ(g)∈YZ$ as $fˉ(g)(z):=f(g(z))(z)$? Seems like a canonical choice to me, although it might fail to be natural in all of the three variables as you say.

Yes, I think maybe this is a special case of the following? As long as $\mathcal{V}$ has a terminal object $\mathbf{1}$ (as in the case of sets), given $\alpha : R \to \{g,P\}$, naturality of cotensoring gives a map $\{g,R\} \to \{\mathbf{1},R\} \cong R$, which can then be composed with $\alpha$. This would solve my problem, actually. (Edit: this is slightly wrong - I actually don't know that $\{\mathbf{1},R\} \cong R$ unless $\mathbf{1}$ is the monoidal unit.)

But maybe I'm confused about something, since this seems to work for any $g$?

Ari Rosenfield (she/her) (Mar 14 2025 at 15:13):

In case it wasn't clear why I don't expect it to hold for arbitrary $g$, I think this would imply that sheafification preserves some infinite powers of presheaves, and this shouldn't be the case in general.

Could you say more about what compactness assumptions you wanted to make on $g$/why they would let me avoid this problem?

Morgan Rogers (he/him) (Mar 14 2025 at 16:24):

Wait never mind, I had forgotten how weak preservation of powers actually is (or rather, I had forgotten that it was automatic for sheafification). There is a standard result (Proposition 1 in Section III.6 of Sheaves in Geometry and Logic) that if $F$ is a sheaf and $P$ a presheaf then the exponential $F^P$ is a sheaf too. In particular this applies to the case where $P$ is the constant presheaf at some set $g$, when this coincides with $\{g,F\}$. So ignore my complaints.

Ari Rosenfield (she/her) (Mar 14 2025 at 17:23):

Yes, I think maybe this is a special case of the following? As long as $V$ has a terminal object $1$ (as in the case of sets), given $α:R→{g,P}$, naturality of cotensoring gives a map ${g,R}→{1,R}≅R$, which can then be composed with $α$. This would solve my problem, actually.

Oh no, this argument only works when the monoidal unit in $\mathcal{V}$ happens to be terminal :sad: