This may be a fairly simple question. Suppose *G* is a (T0) topological group. Assume that *G* is path-connected, locally path-connected, and semilocally simply connected, so that covering space theory applies.

Question: Is it true that for any element of $\pi_1(G,e)$ (where *e* is the identity element of *G*), there exists a [ADDED: *continuous*] homomorphism from $S^1$ to $G$ having that element of $\pi_1(G,e)$ as its homotopy class?

Another way of formulating this is that there is a set map:

$$\operatorname{Hom}_{cts}(S^1,G) \to \pi_1(G,e)$$

The subscript cts is to indicate continuous.

(when *G* is abelian, the left side has a group structure too [ADDED: *under pointwise multiplication*], and the Eckmann-Hilton principle tells us that we get a group homomorphism).

- Is the set map surjective in all cases (regardless of whether
*G*is abelian)? - Does the image of $\operatorname{Hom}(S^1,G)$ generate $\pi_1(G,e)$ as a group (this is equivalent to surjectivity when $G$ is abelian)?
- Does surjectivity work for Lie groups? Compact Lie groups?
- Does the weaker formulation (2) work for Lie groups?

I have a sketch of an argument/proof that may show (4) (basically, using properties of one-parameter subgroups), but I'm hoping somebody will have a clean proof that works in general for topological groups.

iscompact after all) This would then interact with the monoid structure given by pointwise multiplication, and I'm pretty sure they share the same identity. If interchange holds, then we know that Hom(S^1,G) is an abelian monoid. Whether your map is a homomorphism is an interesting question (to me at least). And as a warning, the proof for (4) will probably fail for Frechet Lie groups ... $\endgroup$ – David Roberts Jan 14 '11 at 22:49