TITLE: Each element of fundamental group of a topological group represented by homomorphism?
QUESTION [13 upvotes]: 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.

REPLY [11 votes]: 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) $S^1\to G$.