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)?


There is a bit more structure in this long exact sequence that you can use. If you have $F \to E \to B$, then on choosing a basepoint in $B$, then for any basepoint $e$ in $F$ you have the sequence

$\dots \to \pi_1 E \to \pi_1 B \to \pi_0 F \to \pi_0 E \to \pi_0 B$.

The extra structure is the fact that $\pi_1 B$ acts on $\pi_0 F$, through deck transformations, if $E\to B$ is actually a covering map, and that:

* the stabilizer group of $[e]$ in $\pi_0F$ under this action is the image of $\pi_1E \to \pi_1B$, and

* the set of orbits $\pi_0F/\pi_1B$ under this action is exactly the "kernel" of $\pi_0E\to \pi_0B$.

Thus, if you have some non-trivial extension and are trying to compute $\pi_1B$, then you get to use the information about deck transformations. I don't know what more to say without an explicit example.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.