1
$\begingroup$

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.

Please help :)

$\endgroup$
5
  • $\begingroup$ 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? $\endgroup$
    – Thibaut Benjamin
    Feb 14, 2020 at 15:07
  • $\begingroup$ No and yes, respectively, @ThibautBenjamin. I have a rough idea how to do the former though. $\endgroup$
    – Shaun
    Feb 14, 2020 at 15:11
  • 2
    $\begingroup$ 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 $\endgroup$
    – Thibaut Benjamin
    Feb 14, 2020 at 15:21
  • $\begingroup$ @ThibautBenjamin it seems clear from the problem statement that an infinite product will be needed. $\endgroup$
    – Kevin Arlin
    Feb 14, 2020 at 16:22
  • $\begingroup$ @KevinCarlson Indeed, I read "finite limits" instead of "infinite limits" $\endgroup$
    – Thibaut Benjamin
    Feb 14, 2020 at 16:26

1 Answer 1

Reset to default
2
$\begingroup$

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?

$\endgroup$
2
  • 1
    $\begingroup$ Does this means then that the forgetful functor is not right adjoint to a "free continous G-set" functor ? $\endgroup$
    – jeanmfischer
    Feb 14, 2020 at 17:04
  • 2
    $\begingroup$ @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. $\endgroup$
    – Kevin Arlin
    Feb 14, 2020 at 17:06

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .