# $S^{2n}$ is the universal cover of $B$, what is $\pi_1(B)$.

Some students and I have tried to solve this problem in the following ways:

1. Using degree theory and results about deck transformations.

2. Using that $S^{2n}$ is the covering space of $\mathbf{RP}^{2n}$ and trying to see if $\mathbf{RP}^{2n}$ must cover $B$ or vice versa.

3. The question was part two of a three part question. Part one asked to show that $\chi(E)=k\chi(B)$ ($\chi$ being Euler characteristic) when $E\to B$ is an $k$-sheeted covering space between $CW$-complexes. We do not know the number of sheets of the cover in the problem, or that $B$ is even a $CW$-complex. If we knew that the number of sheets was finite, and that we could push a $CW$-structure down to $B$, then the result could be used. Intuitively, it seems like $S^{2n}$ cannot be an infinite sheeted cover of any space but we're having trouble proving it.

Any insight into which of these methods is worth putting more thought into would be really helpful. Personally, I would like to know if method 2 could be used at all. I know in general if two spaces have the same universal cover then one need not cover the other, but what if the universal cover has property $x$?

• In every covering $\pi \colon E \to B$, the fibres $\pi^{-1}(y)$ are discrete. Can you have an infinite discrete subset in $S^{2n}$? Sep 25, 2013 at 16:05
• The hairy ball theorem will be useful here. So too will the fact that universal covers are also regular covers (think deck transformation group). Sep 25, 2013 at 16:19
• I guess the point was that $\chi(S^{2n})=2$, hence either $n=1$ and $\pi_1(B)$ is trivial, or $n=2$ and thus $\pi_1(B)$ is the 2-element group. Sep 25, 2013 at 16:46
• There are too many $n$'s in the question and in @user8268's comment (which I upvoted nevertheless...) Sep 25, 2013 at 16:54
• @GeorgesElencwajg oops, I should have written $ň=1$ or $2$ Sep 25, 2013 at 17:26