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

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.

$$i_G$$ 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.