Let $G$ be a commutative topological group (e.g. $S^1$), and let $BG$ be its classifying space. Since $G$ is commutative, the space $BG$ is a group up to homotopy. It is well-known that we have a natural isomorphism
$$\pi\_0Map\_\*(BG,BG) \cong \pi\_0Hom(G,G)$$
where $Map\_\*(BG,BG)$ is the space of pointed maps $BG \to BG$ with pointwise group structure. Clearly, we have a map $\pi\_0Hom(BG,BG) \to \pi\_0Map\_\*(BG,BG)$ induced by the map 
$$Hom(BG,BG) \to Map\_\*(BG,BG).$$
Is the map on path-components an isomorphism of groups? Even better, is the latter map a homotopy equivalence of spaces?



Are all pointed maps of classifying spaces of abelian groups homotopy equivalent to homomorphisms?
I don't think this is true, even for very simple examples. It is not completely evident that your question is well-defined (i.e., independent of the individual model that is chosen for $BG$), but let me try to describe a counterexample nonetheless.


Let $\alpha \colon K({\bf Z},2) \to K({\bf Z},4)$ be the map of spaces classifying the square of a generator of $H^2(K({\bf Z},2);{\bf Z}) \cong {\bf Z}$, and let $\Omega \alpha \colon K({\bf Z},1) \to K({\bf Z},3)$ denote the corresponding $E\_1$-map that arises from looping $\alpha$. Then the E\_1-map $$K({\bf Z},1) \times K(Z,{\bf 3})\to K({\bf Z},1) \times K(Z,{\bf 3}),(x,y) \mapsto (\*,\Omega\alpha(x))$$ is nullhomotopic as a map of spaces, but not as a E\_1-map, since otherwise $\alpha$ would be null.


Thus, for $BG = K({\bf Z},1) \times K({\bf Z},3)$, the map you wrote down fails to be injective on $\pi\_0$.


