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I am trying to understand what the classifying space functor $G \mapsto BG$ does to group homomorphisms $f : H \to G$ for $H, G \in \textbf{TopGrp}$. Is there any suggested reference on the construction of $Bf$?

In particular, let's take the easy example $\mathbb{Z}$ and $\mathbb{Z}_2$. There are two valid homomorphisms between these groups: $n \mapsto [0]$ and $n \mapsto [n \text{ mod } 2]$. Under the classifying space functor, what can I say about the image of these two homomorphisms? (Note that $B\mathbb{Z} \cong S^1 $ and $B\mathbb{Z}_2 \cong \mathbb{R}P^{\infty}$, which may provide at least some geometric intuition.)

My question is a special case of this unanswered one. As also stated there, $Bf$ is the classifying map for the G-bundle $(EH×G)/H→BH$. Do help me see what these are in my example.

Ben Steffan
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Arnav Das
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  • Let's start with your model for $BG$! How do you prefer to construct this space for an arbitrary discrete group $G$? – diracdeltafunk Oct 22 '24 at 00:37
  • @diracdeltafunk For this example, I am inclined to resort to $B\mathbb{Z}_2 \cong BO(1) := Gr_1(\mathbb{R}^{\infty})$ and the corresponding $EO(1) := V_1(\mathbb{R}^{\infty})$. However, on this note, I am trying to get familiar with any models which are particularly nice to deal with BSpin(n), BSO(n), BO(n) and all morphism spaces therein, so I am happy to defer to any wisdom. – Arnav Das Oct 24 '24 at 18:35

2 Answers2

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For the particular case of the map $f: \mathbb{Z}\to\mathbb{Z}_2$ you mentioned, the above answer of @diracdeltafunk is nice. I will answer you about the first part.

There are different ways of constructing classifying spaces $BG$. There is a categorical construction of classifying space as follows:

Given a category $\mathscr{C}$ we can get a simplicial set out of it. Define $N\mathscr{C}$ be the simplicial set so that, $(N\mathscr{C})_k$ is the collections of $\{A_0,\cdots,A_k\} \subset \text{Obj}(\mathscr{C})$ such that there is a morphism between $A_i \to A_{i+1}$. The face maps and the degeneracy maps are the natural one. This simplicial set is called Nerve of the category.

There is a natural functor $|\bullet|: \textbf{sSet} \to \text{CW}$ which is called geometric realization of the simplicial set. For the category $\mathscr{C}$ we call $|{N \mathscr{C}}|=: B \mathscr{C}$ the classifying space of the category $\mathscr{C}$.

Let, $G$ be a topological group. Now consider a category $G$ which has only one object and the morphisms are given by elements of group $G$. The classifying space for this category is the classifying space $BG$ we usually deal with. Now for a group morphism $f: G \to H$ we have a functor $f$ from $G\to H$, by the naturality of the construction of Nerve and the geometric realization functor we have a map $Bf: BG\to BH$. This easily gives you a construction for the model $Bf$.

[The above construction is called Bar construction, you can find more explicit description of $Bf$ here.]

Perhaps it's difficult to visualize $Bf$ from the above construction. Here is an alternate construction using principal $G$-bundles and their correspondence. In general it's difficult to see how $Bf$ acts on $BG$ if you don't have a good model. There are some exceptions.

If $H$ is an admissible subgroup of $G$ then we have a SES: $$H\xrightarrow{i}G \xrightarrow{\rho} G/H$$ the related $Bi$ and $B\rho$ are little easier to understand as you can see in the above answer. Also, these $Bi$ gives you some very useful fiber sequences such as : $$G/H \to BH \to BG \\ BH\xrightarrow{Bi}BG \xrightarrow{B\rho} B(G/H)$$

Trishan Mondal
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  • Thank you for providing the references. Firstly, could you expand on how the first fibre sequence is valid? It implies that every $BH$ projects ever $BG$ as a $G/H$-bundle when I would have expected injection, $H \subset G \implies BH \hookrightarrow BG$? Secondly, for the G-bundle and/or bar construction (assume $H$ is not subgroup of $G$ now), are the classifying spaces and maps specified explicitly to the extent that replacing some homotopy equivalent space for either the domain or codomain in $Bi : BH \to BG$, or using a different, homotopic map in place of $Bi$, may actually break things? – Arnav Das Oct 25 '24 at 21:37
  • The above questions are motivated by the example $Bi : BSO(1) \to BO(1)$. Knowing that $BSO(1) \cong \ast$, I expected this map to be just a point embedding, but several references treat this as the homotopy equivalent $S^{\infty}$, so that $S^{\infty} \to \mathbb{RP}^{\infty}$ is a double cover instead of a point embedding. Are these two equivalent as classifying maps, and if so, how would you say that? (A refined way to phrase this question is ask whether once we pick a model, we are working in $\textbf{Top}$ or $\textbf{hTop}$, or if it depends on the construction.) – Arnav Das Oct 25 '24 at 21:57
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    For the first question: The condition I have given is $H \subset G$ an admissible subgroup (i.e. $G \to G/H$ is a principal $H$-bundle) so for a model of $Bi$ we can take the bundle $EG \times_{G} G/H \to BG$ and now we prove that the fiber of this is weakly equivalent to $G/H$. – Trishan Mondal Oct 26 '24 at 15:49
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    For the second question: As you can see in the first case we have replaced the $Bi$ (that we get from Bar construction) to another weakly equivalent model. So it did not break things in the first case, but I am not sure for the general case. – Trishan Mondal Oct 26 '24 at 15:58
  • I follow your first response but not your second. What are you referring to as "the first case"? Could you take another look at my second question about how the original and replaced classifying maps are related, i.e. define what is a "weakly equivalent model" in your answer? (Since the domain or targets might change, it doesn't seem clear.) – Arnav Das Oct 27 '24 at 10:57
  • I think your description of $BG$ only works for discrete groups. For general topological groups, one should see the nerve as a simplicial space and take its geometric realization as a simplicial space, see e.g. this nlab page. – Derived Cats Feb 14 '25 at 17:40
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If you fix a "good" model for $BG$, you can describe the map $Bf$ exactly. Without this, we can only describe the map $Bf$ up to homotopy, since $BG$ is defined by a universal property in the homotopy category.

So, I'll describe the two homotopy classes of maps $S^1 \to \mathbb{R}P^\infty$ coming from the two possible group homomorphisms $\mathbb{Z} \to \mathbb{Z}/2$. It's worth pointing out that homotopy classes of maps $S^1 \to \mathbb{R}P^\infty$ are in bijection with elements of $\pi_1(\mathbb{R}P^\infty) \cong \mathbb{Z}/2$. So, we might expect that the trivial homomorphism corresponds to the trivial element of $\pi_1$ (i.e. a nullhomotopic map $S^1 \to \mathbb{R}P^\infty$), and that the nontrivial homomorphism corresponds to the nontrivial element of $\pi_1$. This is true!

In this case, $E\mathbb{Z} = \mathbb{R}$, so the map $S^1 \to \mathbb{R}P^\infty$ that we want is the classifying map of the $\mathbb{Z}/2$-bundle $$\mathbb{R} \times_{\mathbb{Z}} \mathbb{Z}/2 \to S^1.$$

For the trivial homomorphism, the total space is homeomorphic to $S^1 \times \mathbb{Z}/2$ (why?) and the bundle $S^1 \times \mathbb{Z}/2 \to S^1$ is the trivial bundle. Thus, the desired map $S^1 \to \mathbb{R}P^\infty$ is nullhomotopic, as predicted. One possible model is a constant map.

For the nontrivial homomorphism, the total space is $S^1$ (why?), and in particular connected, so the bundle $S^1 \to S^1$ is not trivial. Thus, the desired map $S^1 \to \mathbb{R}P^\infty$ is not nullhomotopic. One possible model is the image of a "half-equator" of $S^\infty$ via the quotient $S^\infty \to \mathbb{R}P^\infty$.

diracdeltafunk
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  • Thank you for the answer, I see that for the trivial homomorphism, the total space is the quotient space under the relation $(r+n,i) \sim (r,i),r∈\mathbb{R},i∈\mathbb{Z}2$, but for the nontrivial homomorphism $\rho_2: n \mapsto [n]_2$, I can't quite see the isomorphism $\mathbb{R}\times{\rho_2}\mathbb{Z}_2\cong S^1$. I get for n even $(r+n, i) \sim (r, i)$ and n odd $(r+n, i') \sim (r+n, i)$, where $i'$ is the element that is not $i$, but I don't see how these cases merge to provide the isomorphism. – Arnav Das Oct 24 '24 at 18:37
  • The identification $(r+n,i) \sim (r,i)$ for $n$ even turns the two copies of $\mathbb{R}$ into two circles $\mathbb{R}/2\mathbb{Z} \cong S^1$. Then the identification $(r+1,i) \sim (r,i')$ identifies the two circles via the antipodal map. We end up with one circle! – diracdeltafunk Oct 26 '24 at 03:08