This is Exercise I.6(a) of Mac Lane and Moerdijk's, "Sheaves in Geometry and Logic [. . .]". According to this search, it is new to MSE.
The Question:
Let $G$ be a topological group and $\mathbf{B}G$ the category of continuous $G$-sets${}^\dagger$. 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${}^\dagger$, iff for each $x\in X$ its isotropy subgroup $$I_x=\{ g\in G\mid x\cdot g=x\}$$ is an open subgroup of $G$.
(I'm not sure what an open subgroup is, unless it's just a subgroup $H\le G$ such that $H$, when considered as $H\subseteq G$, is open in the topology of $G$; I couldn't find a definition)
Thoughts:
Let $G$ be a topological group with topology $\tau$, and $(X,\mu: X\times G\to X)$ a $G$-set.
$(\Leftarrow)$
Suppose, for all $x\in X$, that $I_x$ is an open subgroup of $G$. Then, since $\mu$ is a right $G$-action, we have, for each $g'\in G$, that $(x\cdot g)\cdot_\mu g'=x\cdot_\mu g'=m$ for some $m\in X$ dependent on $x$ and $g'$.
I don't know how to proceed from here.
$(\Rightarrow)$
Assume $\mu$ is continuous. Then, for any open $U\subseteq G$ with respect to the discrete topology, we have $\mu^{-1}(U)$ in the topology $\tau$. But every subset of $G$ is open in the discrete topology; in particular, for each $x\in X$, $\mu^{-1}(I_x)$ is in $\tau$.
(See $\dagger$ below.)
I'm not sure what to do from here on out.
Please help 🙂
$\dagger$: What is the topology on $X$? Perhaps this will make explicit what is meant by "$\mu$ is continuous".
Best Answer
Let $G^{\delta}$ be the topological group $G$ but with the discrete topology. There is a continous group morphism $\iota_G : G^{\delta} \to G$.
Consider a topological group $H$, then denote by $H-\text{Cont}$, the category of continous left $H$-actions, i.e. spaces $X$ and $\mu : H \times X\to X$ continous action. Consider $f:H\to K$ a morphism of topologial groups, there is a functor $$ f^* : K-\text{Cont} \to H-\text{Cont} $$
The morphism $\iota_G$, gives then a functor $$ (\iota_G)^* : G-\text{Cont} \to (G^\delta)-\text{Cont} $$ Given a $G$-set, $(X, G\times X \to X)$ (where here the action is just a map of sets, no topology of $G$ involved) we can form a $G^\delta$-space if we give $X$ the discrete topology. Modulo verifications on what happens to morphisms, there is a functor $$ \Delta : G-\text{Set} \to (G^\delta)-\text{Cont} $$ One interpretation of your question is :
If for all $x\in X$, $I_x$ are open then $\mu^{-1}(\{x\}) = I_x\times \{x\}$ which is open in the product topology of $G\times X$, where $X$ has the discrete topology. So we have an object of $G-\text{Cont}$, and this object is sent to $\Delta(X)$ by the functor $(\iota_G)^*$.
On the other hand suppose that $\Delta(X)$ is in the image of $(\iota_G)^*$, there is some space $Y$ and a continuous $G$-action $\nu : G\times Y \to Y$, such that $(\iota_G)^*(Y) = \Delta(X)$. Since $(\iota_G)^*$ doesn't change the underlying space and the underlying action, $Y$ should be $X$ with the discrete topology and $\nu$ is $\mu$, then $\mu^{-1}(\{x\}) = I_x\times \{x\}$ should be open for every $x\in X$, so $I_x$ is open in $G$.
Another question is the following
Let $X$ be a space and $\mu : G^\delta \times X \to X$ a continous action on $X$. The question is equivalent to : under what conditions is $\mu$ continous map $G\times X\to X$, with $G$ having the topology we started with ?
Define for all $U$ open in $X$, $I_U = \{g\in G, \forall x \in U, g\cdot x \in U\}$. I think is not to difficult now to see that $(X,\mu)$ is in the image of $(\iota_G)^*$ if and only if, for all $U\subseteq X$ open, $I_U$ is open in $G$.