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Stream: learning: questions

Topic: Coequalizers in the Category of G-Spaces

Ruby Khondaker (she/her) (Jul 28 2026 at 14:37):

Let $G$ be a topological group, and $G\mathbf{Top}$ denote the category of $G$-spaces - meaning, topological spaces equipped with a continuous $G$-action. This is exactly the category of algebras for the monad $G \times (-)$ on $\mathbf{Top}$, so the forgetful functor $G \mathbf{Top} \to \mathbf{Top}$ creates all limits, and creates all colimits that the monad and its square preserve.

So $G\mathbf{Top}$ is complete. Since products distribute over coproducts in $\mathbf{Top}$, it also has coproducts that are created by the forgetful functor to $\mathbf{Top}$. I was interested in understanding when it has coequalizers, and what they look like.

If $G$ is locally compact (or core-compact, but this turns out to be equivalent for topological groups), $G \times (-)$ preserves quotient maps. Hence $G\mathbf{Top}$ in this case would be cocomplete and the forgetful functor creates coequalizers.

However, if $G$ is not assumed to be locally compact, I think $G \mathbf{Top}$ can still be argued to have coequalizers. Here is the approach I have in mind:

Let $X, Y$ be $G$-spaces, and $f, g : X \to Y$ be two $G$-equivariant maps. We can define a copresheaf $F : G \mathbf{Top} \to \mathbf{Set}$ where $F(Z)$ is defined to be the set of $G$-equivariant maps $Y \to Z$ that coequalize $f, g$. We seek a representation of $F$, which is equivalently an initial object of $\int F$, its category of elements.

Now, $F$ preserves limits (indeed this is true in any category), so the projection $\pi : \int F \to G\mathbf{Top}$ creates limits. Hence $\int F$ is complete.

It suffices to show that $\int F$ has a weakly initial set of objects. For this, I claim that the collection of objects $(S, \beta)$ where $S$ is a $G$-space whose underlying set is a subset of $Y$ suffices. This forms a set by standard arguments, and any equivariant map $\alpha : Y \to Z$ can be factored as $Y \to \alpha(Y) \to Z$, whence $\alpha(Y)$ is isomorphic to such an $S$ and transport of structure can be used to define the requisite map $Y \to S$.

Of course, this is a rather involved construction - I don't really have any idea what coequalizers explicitly look like when $G$ is non-locally-compact! Two questions:
1) Does this argument look correct?
2) I may have done some redundant work by redoing a variant of the standard adjoint functor theorem argument - is there a nice way to recast finding a coequaliser as some kind of adjoint such that this theorem can be applied?

Morgan Rogers (he/him) (Jul 28 2026 at 15:22):

Skipping over the copresheaf and category of elements arguments, you can directly consider the collection of regular quotients of $Y$ which are $G$-space homomorphisms and coequalize $f,g$, and your claim is that the poset of such is complete, so that there is a least element in the lattice, which will be the coequalizer.

Considering this statement:

If $G$ is locally compact (or core-compact, but this turns out to be equivalent for topological groups), $G \times (-)$ preserves quotient maps. Hence $G\mathbf{Top}$ in this case would be cocomplete and the forgetful functor creates coequalizers.

if $G$ is not locally compact and $G \times (-)$ doesn't preserve quotients, it's not clear to me that the above will be true; if I take quotients $Y \twoheadrightarrow P$ and $Y \twoheadrightarrow Q$ that are $G$-space homomorphisms, will the quotient in the image factorization of the induced map $Y \to P \times Q$ still be a $G$-space homomorphism? I don't know what a standard example of a non-locally-compact group is to check whether or not it so happens that these limits of quotients are well-behaved.

Ruby Khondaker (she/her) (Jul 28 2026 at 15:24):

Hm, perhaps you could just take the additive group of an infinite-dimensional normed space? This is metrizable with the norm topology, but cannot be locally compact (since the unit ball is norm-compact iff the space is finite-dimensional).

Ruby Khondaker (she/her) (Jul 28 2026 at 15:34):

Oh, I think one thing to bear in mind - the solution set I gave isn't quite just the collection of regular quotients? Essentially, we can decompose any equivariant map $\alpha : Y \to Z$ as $Y \to Y / \text{ker}(\alpha) \to \text{Im}(\alpha) \to Y$, where the first is a quotient map, the second is a continuous bijection, and the third is an inclusion.

However, the continuous bijection need not be a homeomorphism. So for the solution set to work, I can't just take all spaces of the form $Y / \text{ker}(\alpha)$ - instead, these need to be "closed up" under allowing for coarser topologies. This is equivalent to taking spaces of the form $\text{Im}(\alpha)$ I believe, and then I use a transport of structure-argument to show that it suffices to think of $\text{Im}(\alpha)$ as a strict subset of $Y$ but possibly re-topologised.

Morgan Rogers (he/him) (Jul 28 2026 at 15:47):

Surely if $Y \to Y/\mathrm{ker}(\alpha) \to \mathrm{Im}(\alpha)$ coequalizes $f,g$ and the second part is a continuous bijection then the first part must coequalize $f,g$?

Ruby Khondaker (she/her) (Jul 28 2026 at 15:48):

Yes I believe so, because coequalising here is just a set-theoretic condition

Morgan Rogers (he/him) (Jul 28 2026 at 15:49):

So why can't you "just take all spaces of the form $Y/\mathrm{ker}(\alpha)$ "?

Ruby Khondaker (she/her) (Jul 28 2026 at 15:49):

Hm maybe you could then actually, I just didn't do that for my original solution

Morgan Rogers (he/him) (Jul 28 2026 at 15:50):

But my question remains of whether $Y \to Y/\mathrm{ker}(\alpha)$ is automatically a $G$-space homomorphism

Ruby Khondaker (she/her) (Jul 28 2026 at 15:53):

Ah I understand better now - I am not actually sure, part of the issue here of course is that while $Y / \text{ker}(\alpha)$ is a topological space, defining a continuous $G$-action on it is difficult when you don't know if $G$ preserves quotients.

So maybe what I did in my solution is bypass the usage of $Y / \text{ker}(\alpha)$ entirely by just using $\alpha(Y)$? One can take subspaces of $G$-spaces (and thus images of $G$-equivariant maps) with no topological issues, and then $Y \to \alpha(Y)$ is a $G$-space homomorphism.

Morgan Rogers (he/him) (Jul 28 2026 at 15:54):

So the image in $\mathbf{Top}$ might not be a coequalizer, but the coequalizer should still exist!

Ruby Khondaker (she/her) (Jul 28 2026 at 15:56):

Yes that's what I'd expect, since $G \times (-)$ doesn't generally preserve coequalisers.

Essentially I started thinking about this by trying to establish categorical properties of $G\mathbf{Top}$, realising "hey wait coequalisers seem hard since I can't obviously use the construction in $\mathbf{Top}$, do they even exist?", and then coming across this argument. For non-locally compact $G$, I would expect that the forgetful functor $G \mathbf{Top} \to \mathbf{Top}$ wouldn't preserve coequalisers, much less create them.

Morgan Rogers (he/him) (Jul 28 2026 at 15:57):

It seems like you end up constructing it from the coequalizer in $\mathbf{Set}$ more or less... I wonder if there's a slick way to show that?

Ruby Khondaker (she/her) (Jul 28 2026 at 15:59):

Hm the construction is sufficiently involved that I’m not actually sure if the other forgetful functor $G \mathbf{Top} \to G \mathbf{Set}$ preserves coequalisers. The underlying set of the coequaliser I constructed is some horrible subspace of a huge product, after all…

Morgan Rogers (he/him) (Jul 28 2026 at 16:01):

Hopefully we've spelled out enough of the details in discussing it that someone will be able to jump in with a neat result if there is one :sweat_smile:

Kevin Carlson (Jul 28 2026 at 16:36):

Do you want to summarize? You've pinned down some "other" forgetful functor and you want to know whether it's known to preserve coequalizers?

Ruby Khondaker (she/her) (Jul 28 2026 at 16:39):

Yes, so to summarise:
1) The original argument shows that, even for non-locally-compact topological groups $G$, the category of $G$-spaces $G \mathbf{Top}$ is still cocomplete.
2) However, for these $G$ the monad $G \times (-)$ no longer preserves quotient maps, so it is not automatic that the forgetful functor $G \mathbf{Top} \to \mathbf{Top}$ would preserve coequalisers.
3) There is also another natural forgetful functor $G \mathbf{Top} \to G \mathbf{Set}$ - I am not sure whether this preserves coequalisers, either. Since $G$ is a topological group, $G \mathbf{Set}$ is no longer just a functor category, so I'm not even sure what coequalisers in it look like.

My questions are then:
A) Is the argument I presented for "1" correct?
B) Is there a slicker argument available?
C) What do coequalisers in $G \mathbf{Top}$ look like generally, and how do they interact with these forgetful functors?

Kevin Carlson (Jul 28 2026 at 16:48):

Ah, I figured you meant a non-obvious forgetful functor to $G\mathbf{Set},$ thanks.

Ruby Khondaker (she/her) (Jul 28 2026 at 16:49):

Right so since $G$ is a topological group, one would probably define $G \mathbf{Set}$ as sets equipped with a $G$-action $\rho : G \times X \to X$ that is continuous when $X$ is considered a discrete space? Or one could relax the continuity requirement which is equivalent to forgetting the topology on $G$, but I'm not sure if that's sensible.

Kevin Carlson (Jul 28 2026 at 16:52):

Yes, usually $G$-sets for a topological group are defined like that. I think I remember that this just means a $G$-set, ignoring the topology, such that the stabilizers are all open subgroups (Morgan will remember this better), which means that for familiar topological groups that lack open subgroups this is just the category of discrete $G$-sets.

Ruby Khondaker (she/her) (Jul 28 2026 at 17:46):

Actually I am no longer sure about the forgetful functor in 3 - if G x X -> X is continuous, it shouldn’t be generally true that G x Disc(X) -> Disc(X) is continuous, right? Maybe we do need to forget the topology on G then.

Mike Shulman (Jul 28 2026 at 23:38):

Do you have a specific reason to care about $G$-actions in the category of all topological spaces? If you use a [[convenient category of spaces]], then the monad $G\times (-)$ does preserve all colimits since it is a left adjoint.

Mike Shulman (Jul 28 2026 at 23:44):

Anyway, my guess would have been that the forgetful functor you'd want would be to sets with an action by the underlying discrete group of $G$. In particular, that functor has a right adjoint (any action on an indiscrete space is continuous), so it preserves colimits and you could hope that it creates them. Perhaps it is even a [[topological functor]]?

Ruby Khondaker (she/her) (Jul 29 2026 at 00:31):

I’ve just been going through Tom Dieck’s book on algebraic topology - $G \mathbf{Top}$ is mentioned in chapter 1 so I was investigating its categorical properties. I’ve heard of these convenient categories but don’t know their details - I suppose I thought it would be good for me to get practice first with $\mathbf{Top}$ itself.

Mike Shulman (Jul 29 2026 at 00:43):

Convenient categories are actually easier, and almost never do you need to know exactly what the definition of your convenient category is, or even which convenient category you're working in. I think the best way to learn algebraic topology is to just pretend that "Top" is cartesian closed, and if anyone complains, wave your hands and say "convenient category". (-: