What's the topology on the mapping space $Map_H(G, Y)$ when $G$ is not finite

Question

When $G$ is a finite group and $H$ a closed subgroup of it, the sets of right cosets $H\backslash G$ has the discrete topology on it. Let $Y$ be a $H-$space. We have the $G-$homeomorphism \begin{equation}\mbox{Map}_H(G, Y)\cong \prod_{H\backslash G}Y.\end{equation} Fix a family of representatives $\{b_{\tau}\}_{\tau\in H\backslash G}$ of the right cosets of $H$ in $G$. For any $g\in G$, there is a unique $b_{\tau}$ and $h''\in H$ such that $g=h''b_{\tau}$. The homeomorphism is defined by $$f\mapsto (f(b_{\tau}))_{\tau\in H\backslash G}.$$ The left $G-$action on $\prod\limits_{H\backslash G}Y$ is defined by \begin{equation}g\cdot (y_{\tau})_{\tau\in H\backslash G}=(h_{\tau'}y_{\tau'})_{\tau\in H\backslash G},\end{equation} where for each $\tau\in H\backslash G$, there is a unique $\tau'\in H\backslash G$ and unique $h_{\tau'}\in H$ s.t. $Hb_{\tau}g=Hb_{\tau'}$, $b_{\tau}g=h_{\tau'}b_{\tau'}$.

If $G$ is a compact Lie group, $H$ a closed subgroup of it and $Y$ a $H-$space, we still have the injection \begin{equation}\mbox{Map}_H(G, Y)\longrightarrow\prod_{H\backslash G}Y\end{equation} by sending a map $f$ to $$\{f(b_{\tau})\}_{\tau\in H\backslash G}.$$ But it may not be surjective because the topology on $G/H$ may not be discrete.

How can I see the topology on the mapping space $Map_H(G, Y)$? What condition a map $f$ should satisfy regarding the topology? Thanks.

Answer 1

Maybe the following is helpful:

If the $H$-action on $Y$ is trivial, then $Map_H(G,Y)\cong Map(H\backslash G,Y)$.

This generalizes to the following: $G\times_H Y \to H\backslash G$ is a fiber bundle with fiber $Y$ whose space of sections is $Map_H(G,Y)$. Note the special case of trivial $H$-action giving a trivial bundle.