Let $X$ be a Polish space with a complete compatible metric $d$. Let $G$ be the group of all isometries of $<X,d>$ with pointwise convergence topology and the composition operation. Fix a countable dense set $D \subset X$.
Let $Z\subset Y$ be the set of all maps which preserve the metric $d$ and have range dense in $X$. So elements of $Z$ are maps from $D\longrightarrow X$.
Conclude that the map $h:G\longrightarrow Z $ defined by $h(g)=g\upharpoonright D$ is a homeomorphism of the spaces with the respective pointwise convergence topologies.
I have shown that $h$ and $h^{-1}$ are continuous and $h$ is onto. So all that is left to show is that $h$ is one-to-one
Please, help!
I started with:
Let $h(g_1)=h(g_2)$. Then $g_1\upharpoonright D = g_2\upharpoonright D$.
Can I some how say that because of this when we extend the functions to $X\longrightarrow X$, they are still equivalent?
Let $x \in X$. Then, as $D$ is dense in $X$, there is a sequence $(x_n) \in D^{\mathbf N}$ such that $x_n \to x$. As $g_1$ and $g_2$ are continuous on $X$, we have $$ g_1(x) = \lim_n g_1(x_n) = \lim_n h(g_1)(x_n) = \lim_n h(g_2)(x_n) = \lim_n g_2(x_n) = g_2(x) $$ Hence $g_1 = g_2$ if $h(g_1) = h(g_2)$.