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Given a hyperbolic 3-Manifold $M=\Gamma_{0}\setminus\mathbb{H}^3$, and a smooth, connected, compact immersed negatively curved surface $\Sigma=\Gamma\setminus\widetilde\Sigma\subset M$, where $\widetilde\Sigma\subset\mathbb{H}^3$ is the universal covering of $\Sigma$, and provided $\Gamma\leq\Gamma_{0}\in$$PSL(2,\mathbb{C})$.

My question is that if $\widetilde\Sigma$ is an embedding in $\mathbb{H}^3$, or it is just an immersion? Or what conditions on $\Sigma$ guarantee that $\widetilde\Sigma\subset\mathbb{H}^3$ is an embedding? (Obviously, it is an embedding when $\Sigma$ is embedded in $M$. )

Thanks for answering.

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    $\begingroup$ It would be an embedding if and only if $\Sigma$ is an incompressible surface in $M$. $\endgroup$
    – Ryan Budney
    Commented Aug 11, 2015 at 4:53
  • $\begingroup$ It sounds wonderful to me. Could you remind me where I can get a reference? Thank you so much. $\endgroup$
    – Nyima Kao
    Commented Aug 11, 2015 at 14:42
  • $\begingroup$ This page has a few of the equivalent definitions of incompressible surface: en.wikipedia.org/wiki/Incompressible_surface The one using the language of fundamental groups is the one you need for this. With that definition it's just the lifting property of a covering space that gives you the result. $\endgroup$
    – Ryan Budney
    Commented Aug 11, 2015 at 19:57
  • $\begingroup$ Your question is very confusing: You are assuming that $\tilde\Sigma$ is a subset of $H^3$. Do you want to know if the map $\tilde \Sigma\to H^3$ is proper? In general, it is best not to talk about "immersed surfaces" but about "immersions". With this in mind, your question might be interpreted as "Suppose that $f: \tilde \Sigma\to H^3$ is an injective immersion and that $f$ is equivariant with respect to an embedding $\Gamma\to \Gamma_0$. Does it follow that $f$ is an embedding, i.e., is a proper map? Answer to this is definitely positive and follows from equivariance condition and ... $\endgroup$
    – Misha
    Commented Sep 5, 2015 at 17:42
  • $\begingroup$ ...the fact that $\Gamma\to PSL(2,C)$ is a discrete embedding. $\endgroup$
    – Misha
    Commented Sep 5, 2015 at 17:42

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