MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $G$ be a discrete group and let $BG \simeq K(G,1)$ be its classifying space. Let $H$ be a topological group with classifying space $BH$.

  • In case $H$ is also discrete, it was pointed out in the (comments to one) answer of this question that $hom(G,H)$ is in bijection with $[BG,BH]_0$.

Is there anything along the lines of the above statements that can be said in case $H$ is a nonabelian topological group?

Let $A \to C$ be a (topological) crossed module. This also has a classifying space $B(A \to C)$ - the classifying space of the associated $2$-group.

What is the relationship between the first group cohomology of $G$ with coefficients in $A \to C$, i.e. $H^1(G, A \to C)$ and $[BG, B(A \to C)]_0$.

share|cite|improve this question
Dear @Ulrich Pennig: For the case of $H$ a topological abelian group, do you mean to replace $\operatorname{hom}(G,H)$ with its set of path components? Also, do you have a reference for that result? Thanks. – Ricardo Andrade Oct 21 '13 at 10:16
@RicardoAndrade: I have to check what I really meant in the topological abelian case :-) and deleted the statement for now. Sorry for causing confusion. – Ulrich Pennig Oct 21 '13 at 11:50
With regard to the first question, the answer is much more complicated, even for compact Lie groups; see the answers to this question. – Danny Ruberman Oct 21 '13 at 14:08
In some sense, the first question is addressed in a number of my papers. For instance, one can get information about the relationship between Hom($\pi_1 M^g$, U(n)) and Map($M^g$, BU(n)), where $M^g$ is a Riemann surface, using Yang-Mills theory. For sufficiently large n, these spaces have the same homotopy groups above dimension zero. General results on the low-dimensional difference between the homotopy groups of Hom(G, U(n)) and Map(BG, BU(n)) appear in my paper with Baird (arXiv:1206.3341), which also addresses the case of general linear representations. – Dan Ramras Oct 21 '13 at 18:16
@DanRamras: Thanks! I will look into it. – Ulrich Pennig Oct 22 '13 at 5:23

Let $\mathbf{H}$ be an $\infty$-topos (in your case, you want to take the $\infty$-topos of spaces, $\mathcal{S}$). There is an equivalence of $\infty$-categories $$ \Omega : \operatorname{Grp}(\mathbf{H}) \to \mathbf{H}^{\ast/}_{\geq 1}, \quad \mathbf{B} : \mathbf{H}^{\ast/}_{\geq 1} \to \operatorname{Grp}(\mathbf{H}) $$ between the $\infty$-categories of $\infty$-group objects in $\mathbf{H}$ (with $\infty$-group homomorphisms between them) and that of pointed, connected objects in $\mathbf{H}$ (where maps should respect the basepoints), mediated by the usual looping and delooping functors —here $\Omega$ is left adjoint to $\mathbf{B}$. This is all in Lurie's Higher Topos Theory, but I recommend the exposition in Nikolaus, Schreiber and Stevenson's Principal $\infty$-bundles - General Theory.

Discrete groups are $0$-truncated objects in $\operatorname{Grp}(\mathcal{S})$, and the equivalence above implies an equivalence of mapping spaces: $$ B : \operatorname{Map}_{\operatorname{Grp}(\mathcal{S})}(G, H) \to \operatorname{Map}_{\mathcal{S}^{\ast/}_{\geq 1}}(\mathbf{B}G, \mathbf{B}H) $$ The left hand side here is homotopy equivalent to the set of group homomorphisms from $G$ to $H$, and taking $\pi_0$ gets you the desired bijection: $\hom(G, H) \cong [\mathbf{B}G, \mathbf{B}H]_0$.

If $H$ is not discrete, $\mathbf{B}H$ is no longer an Eilenberg-Maclane space: rather, $\pi_{i+1}(\mathbf{B}H, \ast) = \pi_i (H, e)$.

  • If what you really want is $K(H,1)$, you can give $H$ the discrete topology —call it $H^\delta$. Then you have $\mathbf{B}H^\delta \simeq K(H,1)$, and a bijection $$\hom(G, H) \cong \pi_0\operatorname{Map}_{\operatorname{Grp}(\mathcal{S})}(G, H^\delta) \cong [K(G, 1), K(H, 1)]_0$$ Forgetting the topology on $H$ is no big deal because $G$ is discrete.

  • For the classifying space $\mathbf{B}H$ things are more complicated. The most you (or at least I) can say is that there is a bijection $$\pi_0\operatorname{Map}_{\operatorname{Grp}(\mathcal{S})}(G, H) \cong [\mathbf{B}G, \mathbf{B}H]_0$$

As for your second question, the first group cohomology group with values in the 2-group associated to the crossed module $A \to C$ is given by the set of homotopy classes of (unbased) maps from $\mathbf{B}G$ to $\mathbf{B}(A \to C)$: $$ H^1(G, A \to C) = [ \mathbf{B}G, \mathbf{B}(A \to C) ] $$ (see the nLab pages on cohomology and group cohomology).

share|cite|improve this answer
I think that the statement $\operatorname{hom}(G,H) = [BG,BH]_0$ when $G$, $H$ are discrete groups is significantly simpler than the equivalence of homotopy categories between pointed connected spaces and loop spaces (or group objects in the quasi-category of spaces). – Ricardo Andrade Oct 21 '13 at 14:48
@Ricardo: I was just using the discrete case for illustration purposes. I'm sure there are proofs that don't require that much technology. – Alberto García-Raboso Oct 21 '13 at 14:59
@AlbertoGarcía-Raboso: Are you assuming that $A$ and $C$ are again discrete for the last statement? – Ulrich Pennig Oct 21 '13 at 16:13
This is all well, but I was hoping to understand the right hand side, i.e. $[BG,BH]_0$ with the topology on $H$, but maybe the equivalence of mapping spaces is all you get in general. – Ulrich Pennig Oct 21 '13 at 16:17
@Ulrich: in the last statement you do not need to assume that $A$ and $C$ are discrete. – Alberto García-Raboso Oct 21 '13 at 18:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.