# 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 $G \to H$?

• It's not even true for discrete groups unless you go to the pointed homotopy category. – Zhen Lin Sep 30 '15 at 18:52
• @ZhenLin Fair enough, that's what I meant. Let me edit. – user98602 Sep 30 '15 at 18:53

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 $G = S^1, H = SU(2)$.

• Homotopy clases of continuos group homomorphisms are the wrong thing to look at only if they are not what you have to look at. That sort of claim is only true in situations where one has nothing to do but deontics! :-) – Mariano Suárez-Álvarez Oct 1 '15 at 2:58
• @MarianoSuárez-Alvarez: I'm not sure I agree. In any case, I learned something from this answer. – user98602 Oct 1 '15 at 3:32
• @Mike: in general, when $H = U(n)$ this issue is closely related to the difference between Borel-equivariant and genuinely equivariant K-theory (of a point); Borel $G$-equivariant K-theory of a point is the K-theory of $BG$, while genuinely $G$-equivariant K-theory of a point is the representation theory of $G$. These two differ but are related by the Atiyah-Segal completion theorem. – Qiaochu Yuan Oct 1 '15 at 16:52
• Isn't the other summand in $H^2(G,\mathbb{Z})$ given by $\mathrm{Hom}(H_2(G,\mathbb{Z}), \mathbb{Z})$, which vanishes because for a finite group $G$, $H_2(G, \mathbb{Z})$ is finite too? – Omar Antolín-Camarena Apr 22 '16 at 1:33
• In Maps of BZ/p to BG, Dwyer and Wilkerson show that the map Hom(G, SU(2))/conjugation --> [BG, BSU(2)] (that's homotopy classes of maps, not of pointed maps, which is why the domain is conjugation classes of group homomorphisms) is not surjective if G is the symmetric group on 3 letters and also that it is not injective if G is cyclic of order 15. – Omar Antolín-Camarena Apr 22 '16 at 17:46

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.

• How annoyingly trivial. Is this fixable by changing the former category or expanding the latter? (Perhaps by restricting to compact groups?) – user98602 Sep 30 '15 at 19:03
• Well, at minimum you should replace $\mathbf{TopGrp}$ with $\operatorname{Ho} \mathbf{TopGrp}$. But even then I suspect the answer is no. – Zhen Lin Sep 30 '15 at 19:04
• I agree that is a better replacement than restricting to compact groups (though I don't know what the morphisms in your homotopy category are - homotopies through homomorphisms?) Please let me know if you find a contradiction to this statement. – user98602 Sep 30 '15 at 19:10
• Homotopies through homomorphisms is probably the simplest option. We would need to have induced homotopies of the morphisms of bar constructions. – Zhen Lin Sep 30 '15 at 19:40

I know this is an old+answered question, but I don't think the following result has been explicitly brought up:

Let $$G$$ be a Lie group, $$g\in G$$, and let $$c_g\colon G\to G$$ be the automorphism given by $$c_g(h) = ghg^{-1}$$. Then the induced map $$B c_g\colon BG \to BG$$ is homotopic to the identity map.

(see for example Ebert's Notes on index theory, lemma 9.2.2 on page 143)

Then any $$(G, g)$$ such that $$c_g$$ is not homotopic to the identity map (for example $$G$$ discrete and $$g$$ not in the centre) is a counter-example.

Based on Omar Antolín-Camarena's comment on Qiaochu Yuan's answer, it seems like quotienting out the inner automorphisms would not be enough to make it fully-faithful in general either.