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).


1. Is the set map surjective in all cases (regardless of whether *G* is abelian)?
2. 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)?
3. Does surjectivity work for Lie groups? Compact Lie groups?
4. 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.



Each element of fundamental group of a topological group represented by homomorphism?
No. A continuous homomorphism $S^1\to G$ yields a map $BS^1\to BG$. The space $BS^1$ is homotopy equivalent to $\mathbb CP^\infty$. There is a topological group $G$ such that $BG$ is homotopy equivalent to the sphere $S^2$. A map corresponding to a generator of $\pi\_1G=\pi\_2BG=H\_2S^2$ would give an isomorphism $H^2BG\to H^2BS^1$, but this is incompatible with the cup product.


EDIT: This example is universal in the following sense: A standard way of making a Kan loop group for the suspension of a based simplicial set $K$ is to apply (levelwise) the free group functor from based sets to groups. The realization of this is then the universal example of a topological group $G$ equipped with a continuous map $|K|\to G$. Apply this with $K=S^1$.


EDIT: Yes in the Lie group case. It suffices to consider compact $G$ since a maximal compact subgroup is a deformation retract. Now put a Riemannian structure on $G$ that is left and right invariant, and use that the geodesics are the cosets of the $1$-parameter subgroups, and that in a compact Riemannian manifold every loop is freely homotopic to a closed geodesic. 


