Define the lens space L(m,n) as the quotient of S2m+1 by the action of the cyclic group ℤn⊂S1⊂ℂ*. We can create the infinite lens space L(∞,n) by a telescoping construction on the lens spaces L(m,n) for fixed n, which has as an n-sheeted covering space S∞. The homotopy exact sequence is then
... --> π1(S∞) --> π1(L(∞,n)) --> π0(n points) --> π0(S∞) --> ...
Here, π1(S∞) is the trivial group, and π0(S∞) is a set with one point. The sequence is still exact at the π0 portion, once we specify a basepoint for each set and call its preimage the kernel of the map, meaning that whatever π1(L(∞,n)) is, it definitely has n elements. It's a group, too, so if n is prime then we have no choice but to conclude that π1(L(∞,n))=ℤn. Of course, even when n isn't prime, the fact of the matter is that this statement is still true.
But this is a little unsettling to me. It seems like we're only concluding that because the deck group of the universal cover happens to be ℤn (or, admitting the full extent of our complicity, because we're very nearly taking as a definition that L(∞,n)=S∞/ℤn). If we aren't working with the universal cover, then π1 of the cover isn't trivial, so even if the cover is connected it seems like we could run into the extension problem in trying to compute π1 of the base. Of course, this is all algebraic; perhaps there's something geometric that will save the day and tell us how to interpret this. Is that the case, or is there some other way to unambiguously determine π1 of the base here (perhaps from the definition of the connecting homomorphism via the covering homotopy property)?