Let $(X,d)$ be a unique geodesic metric space and $Y\subseteq X$ a subset, equipped with the induced path metric $d'$. We say that the inclusion $i\colon Y\hookrightarrow X$ is uniformly proper if $\forall A>0\ \exists B>0$ such that $d(x,y)< A$ implies $d'(x,y)< B$ for every $x,y\in Y$. Now consider the case of a compact hyperbolic surface $S$ uniformly properly embedded in a hyperbolic $3$manifold $N$ with isomorphic fundamental group $G=\pi_1(S)\cong\pi_1(N)$. I'd like to prove that in this situation the inclusion $j\colon \tilde{S}\to\tilde{N}$ between the correspond universal covers is still uniformly proper. This is my idea: since $S$ is compact, then $G$ is finitely generated and its Cayley graph may be embedded in $\mathbb{H}^2$ via a map $\varphi\colon \mathcal{G}_G\to\mathbb{H}^2$ such that $\mathcal{G}_G(G)$ is quasiisometric to $\mathbb{H}^2$. Now, if $S$ were not uniformly properly embedded in $N$, there would exist a positive constant $A$ and a sequence $(x_i,y_i)_i\subseteq S\times S$ such that $d_N (x_i,y_i)< A$ but $d_S (x_i,y_i)\to\infty$. Thus, in particular, there exists a sequence $( g_i, h_i ) \subseteq G \times G $ such that $d_{\mathbb{H}^2}\(g_i,h_i\) \to \infty$, i.e. for every $\gamma\in G$ there exists a sequence $k^\gamma_i\subseteq G$, where $k^\gamma_i=\gamma g_i^{1}h_i$ such that $k^\gamma_i$ eventually exits any bounded ball centred at $\gamma$. What I'd like to do is to use this argument to prove my claim, but I don't understand how this should work in the opposite direction. Suppose that $\tilde{S}\hookrightarrow\tilde{N}$ is not uniformly proper. Then there exists a sequence of points such that the distance in $\tilde{S}$ goes to infinity whereas it is bounded in $\tilde{N}$. And then? How can I say that this property actually passes to the quotient? I think there should be a way of using an argument similar to the one above... Thanks in advance!
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