TITLE: Is the space of $G$-maps $G/H \to X$ naturally homeomorphic to $X^H$?
QUESTION [5 upvotes]: Let $X$ be a $G$-space, where $G$ is a (discrete) group. For a subgroup $H$ of $G$, define$$X^H = \{x : hx = x \text{ for all }h \in H\} \subset X;$$$X^H$ is the $H$-fixed point subspace of $X$. Topologize the set of functions $G/H \to X$ as the product of copies of $X$ indexed on the elements of $G/H$, and give the set of $G$-maps $G/H \to X$ the subspace topology. Is the space of $G$-maps $G/H \to X$ naturally homeomorphic to $X^H$?

REPLY [4 votes]: The answer is Yes.
By definition, $Maps(G/H,X)=\prod_{gH\in G/H}X_{gH}$. For define map $\phi:X^H\to Maps(G/H,X)$, let $gH$-th coordinate of $\phi(x)$ be $gx$. We see that $\phi$ is continuous inclusion (because fo all the compositions $X^H\to Maps(G/H,X)\to X_{gH}$ are continuous). If we take arbitrary $G$-map $f\in Maps(G/H,X)$ and $f(eH)=y$, it is easy to see that $y\in X^H$ and $f=\phi(y)$.
Consider the projection $p_e$ of $\prod_{gH\in G/H}X_{gH}$ to the $X_{eH}$. By definition of topology of a product, $p_e$ will be continuous, and $p_e\circ\phi=Id_{X^H}$, so $X^H$ and space of $G$-maps have the same topology.