Given a topological group $G$, we can form its classifying space $BG$; suppose we have chosen some specific construction, say the bar construction. $B$ is a functor – given any homomorphism $G \to H$, it induces a continuous map $BG \to BH$.
For discrete groups $G$ and $H$, these are all the maps $BG \to BH$ up to (basepointed) homotopy; one can prove more or less by hand that any map between Eilenberg MacLane spaces $K(G,n) \to K(H,n)$ are classified up to homotopy by the homomorphisms $G \to H$ they induce on $\pi_n$, and all homomorphisms are induced this way.
Is this true for arbitrary topological groups $G$ and $H$? Do different homomorphisms $G \to H$ induce homotopically distinct maps $BG \to BH$, and are all maps $BG \to BH$ homotopic to a map induced by a homomorphism? Put another way, is the functor $B: \mathsf{TopGrp} \to \mathsf{hTop}_*$ fully faithful?
Because maps into $BG$ represent $G$-bundles, one reason to care is that this would tell us that the only natural way to go from $G$-bundles to $H$-bundles is via a homomorphism $G \to H$; one might expect this to begin with. If one cares about classifying homotopy classes of maps between certain spaces, it immediately tells you what the maps are between the (infinite) lens spaces and $\Bbb{CP}^\infty$.
Edit: as pointed out by Zhen Lin below, this is trivially false. So let's make some minor modifications to avoid his counterexample: either restrict $G$ and $H$ to be compact (even Lie, maybe? this is the case I care about) groups, or alternately pass to the homotopy category of topological groups, where my morphisms are homotopy (through homomorphisms) classes of continuous homomorphisms; these two ideas should be related by the Iwasawa decomposition, by which a locally compact topological group deformation retracts onto its maximal compact subgroup.
Are homotopy classes of continuous maps $BG \to BH$ the same as homotopy classes of homomorphisms $G \to H$?
Best Answer
The functor $B : \mathbf{TopGrp} \to \operatorname{Ho} \mathbf{Top}_*$ is not fully faithful. Indeed, it does not even preserve non-isomorphy: the multiplicative group $\mathbb{R}^\times$ and the discrete group $\mathbb{Z} / 2 \mathbb{Z}$ have the same classifying space (namely, $\mathbb{R P}^\infty$), so if the functor were fully faithful, $\mathbb{R}^\times$ and $\mathbb{Z} / 2 \mathbb{Z}$ would have to be isomorphic as topological groups – but they are not.