Define the lens space L(m,n) as the quotient of S^{2m+1} by the action of the cyclic group ℤ_{n}⊂S^{1}⊂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

... --> π_{1}(S^{∞}) --> π_{1}(L(∞,n)) --> π_{0}(n points) --> π_{0}(S^{∞}) --> ...

Here, π_{1}(S^{∞}) is the trivial group, and π_{0}(S^{∞}) is a set with one point. The sequence is still exact at the π_{0} portion, once we specify a basepoint for each set and call its preimage the kernel of the map, meaning that whatever π_{1}(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 π_{1}(L(∞,n))=ℤ_{n}. 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 ℤ_{n} (or, admitting the full extent of our complicity, because we're very nearly taking as a *definition* that L(∞,n)=S^{∞}/ℤ_{n}). If we aren't working with the universal cover, then π_{1} 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 π_{1} 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 π_{1} of the base here (perhaps from the definition of the connecting homomorphism via the covering homotopy property)?