Is the classifying space a fully faithful functor?

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 $G \to H$?

Solution 1:

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.

Solution 2:

The correct and invariant statement is that the classifying space functor is an equivalence of $(\infty, 1)$-categories between grouplike $E_1$ spaces (this is a homotopically invariant version of "topological group") and pointed connected spaces (and every time I say "spaces" I mean "weak homotopy types"). I believe it's also true that every grouplike $E_1$ space can be modeled by a topological group.

Trying to strictify this equivalence to a statement about homomorphisms beween topological groups is hopeless in general. The maps corresponding to maps $BG \to BH$ are maps $G \to H$ that respect the group operation up to coherent homotopy, so there's no reason to expect the classifying space functor, as a functor on topological groups and taking values in homotopy categories, to be full.

Explicitly, take $G$ to be discrete and take $H = S^1$. Then $[BG, BH] \cong H^2(G, \mathbb{Z})$, which by the universal coefficient theorem has two summands, namely

$$\text{Ext}^1(H_1(G, \mathbb{Z}), \mathbb{Z})$$

and $\text{Hom}(H_2(G, \mathbb{Z}), \mathbb{Z})$. Group homomorphisms $G \to S^1$ only give you terms in the first summand, via the Ext long exact sequence coming from the short exact sequence

$$0 \to \mathbb{Z} \to \mathbb{R} \to S^1 \to 0$$

giving a long exact sequence with connecting map $\text{Hom}(H_1(G), S^1) \to \text{Ext}^1(H_1(G, \mathbb{Z}), \mathbb{Z})$.

The second summand $\text{Hom}(H_2(G, \mathbb{Z}), \mathbb{Z})$ is related to the nontrivial structure implied by respecting group operations up to coherent homotopy. (This means that $f(g_1 g_2)$ isn't equal to $f(g_1) f(g_2)$, but that there's a homotopy relating them which itself is required to satisfy certain axioms. This homotopy is an embellished version of a $2$-cocycle.)

An earlier version of this answer claimed a counterexample for compact Lie groups (the above example with $G$ finite) that doesn't work; thanks to Omar Antolín-Camarena for pointing this out. I don't know enough about e.g. the homotopy theory of $BSU(2)$ to write down a counterexample for compact Lie groups. I might start by looking at $G = S^1, H = SU(2)$.