# Exercise I.6(b) of "Sheaves in Geometry and Logic [. . .]".

My question on part (a) of this exercise is here. Much of the notation used there is used here.

## The Details:

Definition 1: Given two functors

$$F:\mathbf{X}\to \mathbf{A}\quad G: \mathbf{A}\to \mathbf{X},$$

we say that $$G$$ is right adjoint to $$F$$, written $$F\dashv G$$, when for any $$X\in{\rm Ob}(\mathbf{X})$$ and any $$A\in{\rm Ob}(\mathbf{A})$$, there is a natural bijection between morphisms

$$\frac{X\stackrel{f}{\to}G(A)}{F(X)\stackrel{h}{\to}A},$$

in the sense that each $$f$$, as displayed, uniquely determines $$h$$, and conversely.

For convenience:

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 $$G$$.

## The Question:

(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 $$i_G$$.

## Thoughts:

Let $$G$$ be a topological group with topology $$\tau$$ and $$(X, \mu: X\times G\to X)$$ be a $$\mathbf{B}G^\delta$$-object.

Closure of $$r_G(X)$$ under group action . . .

Let $$\xi\in r_G(X)$$. Then $$I_\xi=\{ g\in G\mid \xi \cdot_\mu g=\xi \}$$ is open with respect to $$\tau$$.

Let $$h\in G$$. Then for $$g\in I_\xi$$, we have $$\xi\cdot_\mu g=\xi$$, so . . . What next?

Do I try & show that $$\mu((\xi, h))\in r_G(X)$$?

I'm not sure how to proceed here. I need to show that $$r_G\circ i_G\stackrel{\sim}{\to}{\rm id}_{\mathbf{B}G}$$ and $$i_G\circ r_G\stackrel{\sim}{\to}{\rm id}_{\mathbf{B}G^\delta}$$ such that

$$\frac{(X,\mu: X\times G\to X)\stackrel{f}{\to}\widetilde{Y}}{(i_G(X),\mu)\stackrel{g}{\to}\hat{Y}},$$

where:

• $$\widetilde{Y}$$ is $$(r_G(Y),$$ (some $$G$$-action on $$r_G(Y)$$ defined by $$\mu$$)),

• $$\hat{Y}$$ is $$(Y,$$ (some $$G$$-action on $$Y$$ defined by $$\mu))$$, and

• $$f$$ determines $$g$$ bijectively.

But I have no clue what I'm doing here.

• I understand now what seemed odd on the other question, you should carefully write the definition of a continous $G$-set. For the first part of the question, yes you have to show just that. This gives clues on how to solve the last part. Commented Feb 11, 2020 at 19:17
• I don't understand the downvote. I've provided ample context. Commented Feb 12, 2020 at 20:17

Suppose $$I_x$$ is open in $$G$$, we want to show that $$I_{xh}$$ is open in $$G$$ for all $$h\in G$$. This follows from the fact that $$I_{xh} = h^{-1} I_x h$$, since conjugation by $$h$$ gives a homeomorphism from $$G$$ to itself. I won't reprove this identity here, since its proof can be found many places online, e.g., here or here. (I should point out that the isotropy group is a synonym for stabilizer subgroup).
This shows that $$X\mapsto r_G(X)$$ is well defined on objects, but we also need that it is well defined on morphisms. Suppose $$f:X\to Y$$ is $$G$$-equivariant. We need to show that $$f(r_G(x))\subseteq r_G(Y)$$. Let $$x\in r_G(X)$$. Then $$I_x\subseteq I_{f(x)}$$, since if $$xg=x$$, then $$f(x)g=f(xg)=f(x)$$. Then since $$I_x$$ is open, and $$I_{f(x)}$$ is a subgroup, we have that $$I_{f(x)}$$ can be written as the union of cosets of $$I_x$$, and is therefore open as well. Thus $$X\mapsto r_G(X)$$ is functorial.
Let $$X$$ be a continuous $$G$$-set. Let $$Y$$ be a $$G^\delta$$-set. We need to show that $$\newcommand\Hom{\operatorname{Hom}}\Hom_{G^\delta}(i_G(X),Y) \simeq \Hom_G(X,r_G(Y)).$$ Since $$r_G(Y)$$ is defined as being a sub-$$G^\delta$$-set of $$Y$$, we have a natural map $$\Hom_G(X,r_G(Y))\to\Hom_{G^\delta}(i_G(X),Y)$$ that sends $$f$$ to the composite map $$X\xrightarrow{f} r_G(Y) \hookrightarrow Y$$. We just need to verify that this is a bijection. It's immediately injective, since we're just including the codomain into a larger set. It's also surjective by what we proved in part 1 to show that $$r_G$$ was defined on morphisms.
That is, we know that $$r_G(i_G(X))=X$$, and we know that for any $$G$$-equivariant morphism of $$G$$-sets, $$f:A\to B$$, we have $$f(r_G(A))\subseteq r_G(B)$$. Apply this to a morphism $$f:i_G(X)\to Y$$. Then we have that $$f(X)=f(r_G(i_G(X))\subseteq r_G(Y)$$. In other words, every $$G$$-equivariant morphism from $$i_G(X)$$ to $$Y$$ factors through $$r_G(Y)$$. But this is precisely what it means for our natural map above to be surjective. $$\blacksquare$$