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Stream: learning: questions

Topic: ✔ Set-toposes are cocomplete?

David Michael Roberts (May 31 2023 at 09:57):

If $\mathcal{E}$ is an elementary topos and $f\colon \mathcal{E}\to \mathbf{Set}$ is a geometric morphism, not necessarily bounded, does $\mathcal{E}$ have arbitrary small coproducts? It is immediate that it has arbitrary small coproducts of objects in the essential image of $f^*$.

I guess we might need to assume $\mathcal{E}$ is locally small, but if we don't need this, I'm not fussed.

Jens Hemelaer (May 31 2023 at 11:08):

Yes, a small family of objects in $\mathcal{E}$ can be interpreted as an object $X$ of $\mathcal{E}/f^*(I)$, where $I$ is the index set. The coproduct is then $\gamma_!(X)$, for $\gamma : \mathcal{E}/f^*(I) \to \mathcal{E}$ the étale geometric morphism associated to $I$.

I think you can do this more generally for colimits/limits indexed by small categories: there is a locally connected geometric morphism $\gamma: [\mathbb{C},\mathcal{E}] \to \mathcal{E}$ for any category $\mathbb{C}$ in $\mathbf{Sets}$ (or in $\mathcal{E}$ even) and then a diagram $X$ in $[\mathbb{C},\mathcal{E}]$ has colimit $\gamma_!(X)$ and limit $\gamma_*(X)$.

Reid Barton (May 31 2023 at 11:34):

It could be interpreted that way, but I don't think it's what David meant.

Jens Hemelaer (May 31 2023 at 11:44):

Do you have another interpretation in mind? I am not suggesting to take $\gamma_!(X)$ and $\gamma_*(X)$ as definitions of colimit and limit, I am claiming that they agree with the usual definitions (the constant diagram functor $\gamma^*$ has left adjoint $\gamma_!$ and right adjoint $\gamma_*$).

Reid Barton (May 31 2023 at 11:47):

Sure: a small family of objects of $\mathcal{E}$ is just a family of objects of $\mathcal{E}$ indexed by a set in the ordinary sense.

Jens Hemelaer (May 31 2023 at 11:51):

Ah, I see! That indeed makes sense. To $\{X_i\}_{i \in I}$ you can associate the map $\bigsqcup_{i \in I} X_i \to I$, but in order to do that you already use coproducts. So then the question remains open...

David Michael Roberts (May 31 2023 at 12:35):

Yes, I mean the non-internal family version.

And now i think about it, every $\mathbf{Set}$-topos is locally small, for formal reasons.

Jens Hemelaer (May 31 2023 at 13:37):

I think Example B.3.1.8(b) in the Elephant gives a counterexample then: if $G$ is a topological group, then there is a topos $\mathbf{Unif}(G)$ of uniformly continuous $G$-sets. There is a geometric morphism $\mathbf{Unif}(G) \to \mathbf{Sets}$ and this geometric morphism is bounded if and only if $\mathbf{Unif}(G)$ has small coproducts, which is the case if and only if $G$ has a smallest open subgroup.

David Michael Roberts (May 31 2023 at 13:40):

I think it's true that we get coproducts indexed by any set of a fixed object, though.

Jens Hemelaer (May 31 2023 at 13:48):

Yes I would say so as well, but I can't think of an argument. Maybe somehow using that products of a fixed object exist, because that is an exponential object?

Morgan Rogers (he/him) (May 31 2023 at 13:55):

David Michael Roberts said:

I think it's true that we get coproducts indexed by any set of a fixed object, though.

Is this "just" the observation that $\coprod_{i \in I} X \cong X \times \coprod_{i \in I} 1$? (that is, the fact that products distribute over any coproducts which exist)?

David Michael Roberts (May 31 2023 at 23:44):

OK, so here's an example: take for $G$ the infinite product $(\mathbb{Z}/2)^{\mathbb{N}}$ with the normal filter of open subgroups given by the finite-index subgroups. A $G$-set has uniformly continuous action precisely when there is an upper bound on the size of its orbits (edit: this is wrong: it's necessary, but not sufficient. You also need the factors sufficiently far down the line, uniformly for all orbits, to act trivially). So one has lots of infinite coproducts, for example finite coproducts of infinite coproducts of the form $\coprod_{i\in I} X$, but not arbitrary coproducts. Since this is locally small the global sections functor from $Unif(G)$ exists and is the direct image part of a geometric morphism to $\mathbf{Set}$. This topos has an nno, inheriting it from $\mathbf{Set}$, I believe.

Notification Bot (May 31 2023 at 23:44):

David Michael Roberts has marked this topic as resolved.

Jens Hemelaer (Jun 02 2023 at 08:04):

As an alternative example, you can also take the elementary topos consisting of $\mathbb{Z}$-sets for which there is an upper bound $n$ on the size of the orbits. These are the ones that are uniformly continuous for the profinite topology (the stabilizers all contain $m\mathbb{Z}$ for $m = n!$).

David Michael Roberts (Jun 02 2023 at 10:07):

Oh, yes. That's even easier :-) the open subgroups are the non-trivial ones!