Fundamental group of a thick part of hyperbolic manifold Let $M$ be a complete hyperbolic manifold of dimension $n$, let $\varepsilon=\varepsilon_n$ be the Margulis constant. Let $M_{[\varepsilon,\infty)}$ be the thick part of $M$ with respect to $\varepsilon$. Suppose that $\pi_1(M)$ is infinite. Is it true that $\pi_1(M_{[\varepsilon,\infty)})$ is also infinite. Or, in case $M_{[\varepsilon,\infty)}$ is not connected, whether there exist a connected component of $M_{[\varepsilon,\infty)}$ such that its fundamental group is infinite.
PS: I am reading a paper, where this fact seems to be an important point in the proof. This is totally not my area, so it might as well be trivial for anyone familiar with these notions, however, I will be much obliged for an expalnation why this is (or is not) true.
 A: Yes, because $\pi_1$ of the boundary of a component of the $\epsilon$-thin part always
surjects to the fundamental group of that thin part, except in one special 2-dimensional
case: if there is a short orientation-reversing curve on a surface, then its component of
the thin part is a Moebius band, and the boundary generates a subgroup of index 2. This
can be dealt with by passing to the orientable double cover, which still would have
infinite fundamental group. 
Except for the mild exception above, $\pi_1(M_{[\epsilon,\infty)})$
maps surjectively to $\pi_1(M)$, so if the latter is infinite the former is infinite.
A: By Margulis lemma, components of the $\varepsilon$-thin part are cusps or $\varepsilon$-tubes, so the interior of the $\varepsilon$-part is $M$ with cusps chopped off, and a finite (possibly empty) collection of embedded closed geodesics deleted. Thus the thick part is connected, and by general position the inclusion of the thick part into $M$ is $\pi_1$-surjective when $n=3$, and is a $\pi_1$-isomorphism when $n>3$. In particular, the fundamental group of the thick part is infinite, because it surjects onto an infinite group $\pi_1(M)$.
EDIT: in the above I assumed that $n>2$. 
