Each element of fundamental group of a topological group represented by homomorphism? 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.
 A: 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. 
A: Here's an abelian counterexample: Let $G$ be the following topological group homotopy equivalent (as a space) to $\mathbb RP^\infty$. It is the realization of the simplicial group that corresponds by Dold-Kan to the chain complex in which the only nontrivial chain group, the first, has order $2$. To put it another way, if a group $A$ is abelian then the usual simplicial model for $BA$ is itself a group (in fact, abelian); let $G$ be $BA$ where $A$ has order $2$.
In this topological group, every nontrivial element has order $2$. Therefore there is no nontrivial homomorphism (continuous or not) from $S^1$.
Note that the reasoning here is very different from what it was in my other example. It is not that there is no homotopically nontrivial map $BS^1\to BG$, and it is not (as in Allen Hatcher's comment) that there is no nontrivial $H$-space map (homomorphism up to homotopy) $S^1\to G$.
