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Maybe I understood what you said: for every sequence $(x_i,y_i)$ of pairs of points in $\tilde{S}$ such that $d_{\tilde{N}}(x_i, y_i)\le A$ we project to the quotient. If $p\colon\tilde{N}\to N$ is the projection map, denote $x'_i=p(x_i)$ and $y'_i=p(y_i)$. Clearly the inequality $d_{N}(x'_i,y'_i)\le A$ still holds, therefore we may use uniform properness to get $d_{S}(x'_i,y'_i)\le B$ for some positive constant $B$. Now we use the cocompactness of the action of $\pi_1(N)$ on the set $\left\{(x,y)\in \tilde{S}\times\tilde{S}\mid d_S(p(x),p(y))\le B \right\}$ and we're done. Is it correct?
Shouldn't we prove that if $d_{\tilde{N}}(x,y) \le A$ then $d_{tilde{S}}(x,y)\le B$? The metric on $\tilde{S}$ is induced by the path metric on $\tilde{N}$ so that $d_{\tilde{S}}(x,y)\ge d_{\tilde{N}}(x,y)$ as a path in $\tilde{S}$ is also a path in $\tilde{N}$ so it's enough to take $A=B$ to get that $d_{\tilde{S}}(x,y)\le A$ implies $d_{\tilde{N}}(x,y)\le B=A$. Am I wrong with that?
distance in $\tilde{N}$ is bounded but their distance in $\tilde{S}$ goes to infinity. But this generates a contradiction since the quotient is compact, so they must be at bounded distance (why? isn't the universal cover of a compact surface isometric to $\mathbb{H}^2$?). It doesn't sound very sound to me, though. I guess I misunderstood. Could you please clarify this?
I understand why the hypothesis is trivially true in this case, but thanks for pointing it out. I'm still confused about the second part: why do we have to consider the set of pairs with $d_S(x,y)< A$? I mean: the inequality holds when we look at the distance in the bigger space, so in $\tilde{N}$, and we are assuming that it doesn't hold in $\tilde{S}$, by contradiction. Perhaps, using your hint, we may say the following: if $\tilde{S}\hookrightarrow \tilde{N}$ were not uniformly proper, than we would have this sequence of pairs of points $(x_i,y_i)$ in $\tilde{S}$ such that their...
@Sam: I will read Thurston's notes! Thanks! @Misha: actually what I was about to do was trying to figure out how to "lift" uniform properness to universal covers. Thank you for the hint!
Thank you everyone for your help. Actually what I meant is what I clearly asked in the first part of my question, i.e. that the inclusion $(S,\rho)\hookrightarrow (N,d_N)$ is uniformly proper. In the last part of the question I wrote something very confused: what I was trying to point out is that the result of Minsky holds for loops while I needed a statement about distance between points, that is length of shortest paths between points. Anyway, your explanation about the diameter of $S$ solved my problems. Thank you very much and sorry about my inaccuracy.
