Let $N$ be an oriented hyperbolic 3manifold of finite volume and let $\Delta \subset N$ be a smooth connected compact subdomain such that the restriction of the injectivity radius function of $N$ to $\Delta$ is larger than some $\varepsilon > 0$ ($\Delta$ could be a thick part of $N$). Is there some finite cover $\Pi\colon \widehat{N} \to N$ such that the injectivity radius function of $\widehat{N}$ restricted to any component of $\Pi^{1}(\Delta)$ is greater than $\varepsilon + 1$?
1 Answer
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As Henry Wilton points out this follows from the fact that $\pi_1 N$ is residually finite, i.e. that for every element $g \in \pi_1 N  \{1\}$ there exist a homomorphism $f_g$ to a finite group $G_g$ such that $f_g: \pi_1 N \rightarrow G_g$ is surjective and $f_g(g)\ne 1 \in G_g$.
Thus the kernel of $f_g$ corresponds to a cover where $g$ and its conjugates do not lift. To complete the proof, we invoke the fact that the set of geodesics shorter than a fixed length (here $\epsilon+1$) is finite.

1$\begingroup$ If you want estimates on the degree you can get a quantitative version of this using the proof of finite residuality. Namely, $\pi_1 N$ embeds in $SL(2,R)$ for some finitely generated integral domain $R$ (in this case it can be taken to be a ring of $S$integers for some finite ser $S$). Then taking the covers corresponding to the morphisms $\pi_1 N \to SL_2(R/nR)$ for noninvertible integers $n \in R$ you get a sequence of manifolds with injectivity radius $\ge \epsilon \log n$ (the volume is of order some power of $n$). $\endgroup$ Commented Jul 14, 2016 at 6:18