Conclude that $\mathbf{B}G$ is a CCC using previous questions in "Sheaves in Geometry and Logic [. . .]".

Question

This is Exercise I.6(c) of the titular book. According the Approach0, this question is new to MSE.

The Setup:

Let $G$ be a topological group and $\mathbf{B}G$ the category of continuous $G$-sets. Let $G^\delta$ be the same group $G$ with the discrete topology. So $\mathbf{B}G^\delta=\mathbf{Sets}^{{G^\delta}^{{\rm op}}}$ is a category as considered in the previous exercise. Let $i_G: \mathbf{B}G\to \mathbf{B}G^\delta$ be the inclusion functor.

(a) Prove that a $G$-set $(X,\mu:X\times G\to X)$ is in the image of $i_G$, i.e., that $\mu$ is continuous, iff for each $x\in X$ its isotropy subgroup $$I_x=\{ g\in G\mid x\cdot g=x\}$$ is an open subgroup of $G$.

Here's my question on part (a).

(b) Prove that, for a $G^\delta$-set $(X,\mu: X\times G\to X)$ as above, the set $r_G(X) = \{x \in X \mid I_x\text{ is open}\}$ is closed under the action by $G$, and that $r_G$ defines a functor $\mathbf{B}G^\delta\to \mathbf{B}G$ which is right adjoint to the inclusion functor $i_G$.

Here's my question on part (b).

The definition of a CCC I am using is as follows.

Definition: A category $\mathbf{C}$ is a cartesian closed category (CCC) if

• it has all finite products (which is equivalent to saying there exists a terminal object and all binary products in $\mathbf{C}$) and

• all $\mathbf{C}$-objects are exponentiable.

The Question:

Observe that $i_G$ preserves products, and conclude from (b) that $\mathbf{B}G$ is cartesian closed since $\mathbf{B}G^\delta$ is. [Hint: define the exponential $Y^X$ in $\mathbf{B}G$ by $r_G(i_G(Y)^{i_G(X)})$.]

Thoughts:

I think I have to make use of the second part of Exercise I.4 but I'm not sure how to implement it.

I have a rough idea of how to conclude that $i_G$ preserves products, and that is to show that it preserves both the terminal object and all binary products. As such, perhaps I can reason as I did for the second part of Exercise I.4, with $F=i_G$. But I doubt this.

Please help :)

Answer 1

$i_G$ preserves finite products

The terminal object in both categories is $\{*\}$ with the unique $G$-action, so the terminal object is preserved. Since $r_G$ is a right adjoint, it preserves all limits. In particular, for all $G^\delta$-sets $X$ and $Y$, $$r_G(X\times_{G^\delta} Y) = r_G(X)\times_G r_G(Y),$$ with the subscripts (somewhat against established convention) denoting the category in which the product occurs.

If $Z$ is a continuous $G$-set, then $r_G(Z)=Z$. Now note that if $X$ and $Y$ are continuous $G$-sets, then $X\times_{G^\delta} Y$ is also a continuous $G$-set, since $(x,y)g = (xg,yg)=(x,y)$ if and only if $xg=x$ and $yg=y$, so $I_{(x,y)} = I_x\cap I_y$. If all the isotropy groups of $X$ and $Y$ are open, then so are all the isotropy groups of $X\times Y$.

Thus if $X$ and $Y$ are continuous $G$-sets, all the $r_G$s in our formula go away, and we get $X\times_{G^\delta} Y = X\times_G Y$, or if we write in the $i_G$s I've been omitting, $$i_G(X)\times_{G^\delta} i_G(Y) = i_G(X\times_G Y).$$

Note that this argument does two things! First it establishes that all of the finite products exist in $\mathbf{B}G$, and second it establishes that the product of continuous $G$-sets is the same as the product in the larger category.

$\mathbf{B}G$ is cartesian closed

We need to show that $-\times Y$ has a right adjoint for all $Y\in\newcommand\BG{\mathbf{B}G}\BG$, which I'll write $[Y,-]$, since I still find exponential notation confusing.

Consider $$\begin{align}\BG(X\times Y, Z)& = \BG(X\times Y, r_G(i_G(Z))) \\ &\simeq \BG^\delta(i_G(X\times Y),i_G(Z)) \\ &\simeq \BG^\delta(i_G(X)\times i_G(Y),i_G(Z)) \\ &\simeq \BG^\delta(i_G(X), [i_G(Y),i_G(Z)])\\ &\simeq \BG(X, r_G([i_G(Y),i_G(Z)])). \end{align}$$

Thus $-\times Y$ is left adjoint to $r_G([i_G(Y),i_G(-)])$, which is what you're asked to show in the question.