# The forgetful functor $U:\mathbf{B}G\to\mathbf{Sets}$ need not preserve infinite limits.

This is Exercise I.7 of Mac Lane and Moerdijk's, "Sheaves in Geometry and Logic [. . .]".

Here $$\mathbf{B}G$$ is the category of all continuous $$G$$-sets, where $$G$$ is a topological group.

## The Question:

Show that the forgetful functor $$U:\mathbf{B}G\to\mathbf{Sets}$$ need not preserve infinite limits.

## Thoughts:

I have considered the contrapositive: that if some $$U':\mathbf{B}G\to\mathbf{Sets}$$ preserves infinite limits, then it is not a forgetful functor. I don't think this is the best way to go about it.

Next, similarly, I thought that I could attempt a proof by contradiction, by assuming $$U$$ always preserves infinite limits, but this time working with the forgetful functor and aiming for a contradiction in $$\mathbf{Sets}$$.

To this end, I have some category $$\mathbf{I}$$ to function as the infinite diagram category that acts on both $$\mathbf{B}G$$ and $$\mathbf{Sets}$$. So consider

\begin{align} U_\mathbf{I}: (\mathbf{B}G)^\mathbf{I} & \longrightarrow\mathbf{Sets}^\mathbf{I}\\ ((X, \sigma)\stackrel{f}{\to}(Y, \tau))^\mathbf{I} & \stackrel{?}{\mapsto} (X\stackrel{f}{\to} Y)^\mathbf{I}. \end{align}

Does the map $$U_\mathbf{I}$$ have to satisfy some universal property or something with respect to $$\mathbf{I}$$? I'm quite lost here.

I'm not sure how to work with $$((X, \sigma)\stackrel{f}{\to}(Y, \tau))^\mathbf{I}$$ and $$(X\stackrel{f}{\to} Y)^\mathbf{I}$$ either, whatever they are.

I think what I have so far might even be nonsense.

• Do you know how to compute any limit explicitly in $\mathbb{B}G$? Have you tried to compute a product of $2$ continuous $G$-sets? – Thibaut Benjamin Feb 14 '20 at 15:07
• Ok, sorry for the obvious question, but you haven't mentionned this in your potential approaches. I would try to compute the product of 2 continuous $G$-sets in the hope that it is not supported by the product of the $2$ underlying sets – Thibaut Benjamin Feb 14 '20 at 15:21
Try to show that $$G$$ generally does not act continuously and consistent with the actions on $$A_i$$ on an infinite product in $$\mathbf{Set}$$ of continuous $$G$$-sets $$A_i$$. Hint: under the action induced from the $$A_i$$, what is the stabilizer of $$(a_i)_{i\in I}$$? What kind of intersections of open subgroups remain open?
• @jeanmfischer That's right. In fact continuous $G$-sets are coreflective in the category of sets with an action of the discrete group $G^\delta$ on the underlying set of $G$. Thus there exists a co-free continuous $G$-set on a set, by composing with the right adjoint to the forgetful functor $G^\delta\to\mathbf{Set}$, but not in general a free one. – Kevin Arlin Feb 14 '20 at 17:06