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Stream: theory: category theory

Topic: ✔ Are topological groupoids a topological concrete category?

Evan Washington (Aug 17 2023 at 18:40):

Hi! I'm trying to get a better feel for what topological groupoids are "like" (and how to construct them). One common way of constructing a topological space is to consider the initial and final topology such that a given family of functions are continuous with respect to that topology. To that end, I have a question: is the category of topological groupoids a topological concrete category over the category of groupoids?

Let $U: \mathsf{TopGrpd} \to \mathsf{Grpd}$ be the functor which takes an internal groupoid in $\mathsf{Top}$ to its underlying internal groupoid in $\mathsf{Set}$ by forgetting the topology. Let $G$ be a groupoid, $(H_i)$ a family of topological groupoids, and $h_i: G \to U(H_i)$ a family of maps in $\mathsf{Grpd}$. The question is, must there be an initial lift? (A topological groupoid $X$ such that $U(X) = G$ with maps $\phi_i: X \to H_i$ such that $U(\phi_i) = h_i$ which is initial among all such)

One reason for thinking this might be true is that the category of topological groups is a topological concrete category over the category of groups. Does this analogy hold, or do we need some 2-categorical generalization (like maybe that the initial lift should be unique only up to equivalence)?

Nota bene: in general, a topological groupoid is not a topological category in the above sense; the names are an unfortunate clash of terminology.

Chris Grossack (they/them) (Aug 19 2023 at 18:37):

Yes!

More generally, fix your favorite essentially algebraic theory $\mathbb{T}$. Then $\mathsf{Mod}(\mathbb{T},\mathsf{Top})$, the models of $\mathbb{T}$ in $\mathsf{Top}$, will be topological over $\mathsf{Mod}(\mathbb{T}, \mathsf{Set})$ via the obvious forgetful functor.

Embarrassingly, I don't actually have a reference for this (though it may be easy to prove? I've never tried.), but it's stated as fact in Section 2.2 of Topological Functors as Total Categories

Chris Grossack (they/them) (Aug 19 2023 at 18:42):

Relatedly, if anyone happens to have a reference that actually proves this, I would be super interested in it. It's almost certainly "do the obvious thing", but I don't have time to work out the details right now. For context, the "obvious thing" is this:

Given a lex functor $F : \mathbb{T} \to \mathsf{Set}$, a family of functors $G_\alpha : \mathbb{T} \to \mathsf{Top}$, and a bunch of natural transformations $F \Rightarrow UG_\alpha$, we want to find a "final lift" $\tilde{F} : \mathbb{T} \to \mathsf{Top}$ so that $U\tilde{F} = F$. We should be able to take the lift "pointwise" in the sense that, at each object $t \in \mathbb{T}$, we get a source $Ft \to UG_\alpha t$, which we can lift to an object $\widetilde{Ft}$. It's not hard to check that this is compatible with arrows in $\mathbb{T}$ as well, so that we can define a functor $\tilde{F}(t) = \widetilde{Ft}$. This is almost certainly the right functor.

But officially we need to check that it's an initial lift of the natural transformations (this should be easy) and moreover that it still preserves finite limits (this seems potentially annoying, but I haven't tried)

Notification Bot (Aug 20 2023 at 02:17):

Evan Washington has marked this topic as resolved.