Let $M$ be a hyperbolic 3-manifold and $\Gamma$ a finitely generated subgroup of $\pi_1(M)$. Then either $\Gamma$ is geometrically finite, or there exists a finite cover $M'$ of $M$ which fibers over the circle, and such that $\Gamma$ is the subgroup of $\pi_1(M')$ corresponding to the fiber ($\Gamma$ is a virtual surface fiber group).
I've seen the result above quoted many times. It is said to be a direct consequence of the Tameness Theorem (every hyperbolic 3-manifold with finitely generated fundamental group is topologically tame) and Canary's Covering Theorem.
I'm having trouble understanding the notion of a geometrically (in)finite end of a 3-manifold (and the definition seems to differ slightly in various places in the literature), and thus I don't understand how Canary's covering theorem leads to the statement above.
I would be very grateful for a rundown of how Canary's theorem implies the subgroup theorem, and also a reference for learning the preliminaries that I need in order to understand Canary's proof.
The precise statement you are after requires $M$ to be of finite volume and complete (otherwise, the claim is false). For simplicity of the discussion, I will assume that $M$ is compact. Here are the steps of the proof:
The quotient manifold $M'=H^3/\Gamma$ has a "compact core", which is a compact submanifold with boundary $C$, whose inclusion in $M'$ is a homotopy-equivalence. The closures of the complementary components $M'-C$ are called "ends" of $M'$ (by an abuse of terminology). There are finitely many of these.
The subgroup $\Gamma$ is geometrically finite if and only if every end is geometrically finite.
Every end of $M'$ is topologically tame (which was known as "tameness conjecture"), i.e. is homeomorphic to the product of a surface and an interval.
If $E$ is a geometrically infinite (topologically tame) end of $M'$, then either $M$ "virtually fibers" and $\pi_1$ of the fiber is a subgroup commensurable to $\pi_1(M')$, or the restriction of the covering map $p: M'\to M$ to the end $E$ is finite-to-one (this is what Canary proves in the linked paper). In fact, Canary actually proves more, namely, that this restriction is a proper map. (One can also formally prove here that finite-to-one implies proper.) Assume that $p|E$ is proper. Since $M$ is compact and $E$ is not, this is a contradiction.
If you want to read more on this staff, consider reading
Matsuzaki, Katsuhiko; Taniguchi, Masahiko, Hyperbolic manifolds and Kleinian groups, Oxford Mathematical Monographs. Oxford: Clarendon Press. ix, 253 p. (1998). ZBL0892.30035.
Ohshika, Ken’ichi, Discrete groups. Transl. from the Japanese by Ken’ichi Ohshika, Translations of Mathematical Monographs. Iwanami Series in Modern Mathematics. 207. Providence, RI: American Mathematical Society (AMS). x, 193 p. (2002). ZBL1006.20031.
Kapovich, Michael, Hyperbolic manifolds and discrete groups, Modern Birkhäuser Classics. Boston, MA: Birkhäuser (ISBN 978-0-8176-4912-8/pbk; 978-0-8176-4913-5/ebook). xxvii, 467 p. (2009). ZBL1180.57001.