Does the sphere $S^3$ nontrivially cover itself?

Question

I am having a hard time deciding whether or not $S^3$ nontrivially covers itself. Some help would be appreciated.

Thanks

Hint: covers are uniquely determined by the induced map on fundamental groups. — Cheerful Parsnip, May 10 '13 at 18:27
$S^3$ is simply connected. What do you know about covering spaces of a simply connected manifold? — 23rd, May 10 '13 at 18:28

score 9 · Answer 1 · answered May 10 '13 at 18:38

We can use Theorem 1.38 from Hatcher's Algebraic Topology to help us here. The theorem essentially says that if $X$ is a path-connected, locally path-connected, and semi-locally simply connected space, then there is a correspondence between the connected covering spaces of $X$ and the conjugacy classes of subgroups of $\pi_1(X)$.

We can check that $S^3$ satisfies the hypotheses of this theorem. As $\pi_1(S^3)$ is trivial (this should be proven) we can conclude that there are no non-trivial coverings of $S^3$.

Does the sphere $S^3$ nontrivially cover itself?

1 Answers1