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 $\mu((\xi, h))\in r_G(X)$?
Right adjoint . . .
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:
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$\widetilde{Y}$ is $(r_G(Y),$ (some $G$-action on $r_G(Y)$ defined by $\mu$)),
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$\hat{Y}$ is $(Y,$ (some $G$-action on $Y$ defined by $\mu))$, and
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$f$ determines $g$ bijectively.
But I have no clue what I'm doing here.
Please help 🙂
Best Answer
First part:
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.
The Adjunction
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. $\blacksquare$