The classifying space of the integer group $\mathbb{Z}$ can be defined as the geometric realization of the underlying groupoid $\mathcal{B}\mathbb{Z}$.

To unwind, $\mathcal{B}\mathbb{Z}$ is simply the category with one object, with $\mathbb{Z}$ as its morphism space. The geometric realization $|\mathcal{B} \mathbb{Z}|$ is a topological space (CW-complex) defined inductively:

  1. To each object, assign a point.
  2. To each morphism, assign a 1-disk (segment) with corresponding end points.
  3. To each 2-tuple of composable morphisms $(f,g)$, assign a 2-disk with corresponding 1-disks as boundary.
  4. To each 3-tuple of composable morphisms $(f,g,h)$, assign a 3-disk... and so on.

It seems like a huge topological space, but there's a theorem I heard several times stating that it's really the classifying space of principal $\mathbb{Z}$-bundles, which turns out to be the $1$-sphere $S^1$ homotopically.

My ultimate goal is to understand (theoretically and pictorially) the proof of the general statement above, for general groups $G$ instead of just $\mathbb{Z}$. I think it will be good to start with this simplest case.

Question: However, I find it rough to visualize how the infinitely defined space above is homotopic to $S^1$. Could you point out how?

More fun: think in this vein for that $|\mathcal{B}\mathbb{Z_2}|$ is $RP^\infty$ and that $|\mathcal{J}|$ is $E\mathbb{Z}_2$, where $\mathcal{J}$ is the unique category with two objects $X,Y$ and four morphisms $f:X\to Y, g:Y\to X$.


1 Answer 1


Since the homotopy type of spaces is determined by their homotopy groups, at least on a convenient category of spaces, it suffices to show that the homotopy groups of $\mathcal{B}\mathbb{Z}$ coincide with those of $S^1$, that is: $$ \pi_n(\mathcal{B}\mathbb{Z}) = \begin{cases} \mathbb{Z} &&\text{ if } n = 1 \\ 0 &&\text { if } n\neq 1 \end{cases} $$

This is equivalent to showing that $\mathcal{B}\mathbb{Z}$ is the first Eilenberg-MacLane space $K(1,\mathbb{Z})$ are weakly homotopic, and in fact this is true for any discrete group $G$. After a couple well known lemmas, the proof outlined below becomes elementary.

First notice that for a discrete group $G$ we have $$ \pi_n(G) = \begin{cases} G &&\text{ if } n = 0 \\ 0 &&\text { if } n > 0 \end{cases}. $$

Recall that $[S^n,X]\cong [S^{n-1},\Omega X]$, and that $\mathcal{B}$ is a delooping i.e. $\Omega\mathcal{B} G\cong G$.

Proposition: If $G$ is discrete then $\mathcal{B}G$ is the first Eilenberg-Maclne $K(1,\mathbb{Z})$.


For $n =1$, \begin{align*} \pi_1(\mathcal{B}G) &= [S^1,\mathcal{B}G] \\ &\cong [S^0,\Omega \mathcal{B}G] \\ &\cong [S^0,G] \\ &\cong \pi_0(G)= G. \end{align*} For $n\geq 2$, \begin{align*} \pi_n(\mathcal{B}G) &= [S^n,\mathcal{B}G] \\ &\cong [S^{n-1},\Omega \mathcal{B}G] \\ &\cong [S^{n-1},G] \\ &\cong \pi_{i-1}(G) = 0. \\ \end{align*} $$ \hspace{15cm}\blacksquare $$

It is interesting to think of what could happen if $\mathcal{B}$ could be iterated i.e. $\mathcal{B}^nG$ made sense. The condition is that $\mathcal{B}G$ is itself a group; surprisingly, for abelian $G$ that can be done and $\mathcal{B}G$ is abelian. For abelian and discrete $G$, such as $\mathbb{Z}$, one could reason exactly as above and show that $K(n,G)\cong \mathcal{B}^n G$. Particularly, we have $\mathcal{B}^n \mathbb{Z}\cong S^n$.

For your More fun question, notice that the question now boils down to showing that $\pi_1(\mathbb{RP}^\infty) = \mathbb{Z}_2$ and all higher homotopy groups vanish. That follows from its orientable double cover $S^\infty$ being contractible.

I have written some account of these facts on a set of lecture notes written for a short course I taught earlier this year; see Section 3. You can also find a lot insight at this blog post by Baez.

  • 1
    $\begingroup$ Thank you for the answer! 1. From this post, homotopy groups themselves do not determine homotopy types.. perhaps in our special case it is. Would you mind elaborating? 2. To use the adjunction, we have to show that the loop of deloop is identity.. but that's essentially my question: how to visualize this fact by my (nerve) construction? (Off topic: I like the Alg->Phys picture in the link you provide!) $\endgroup$
    – Student
    May 6, 2020 at 1:47
  • $\begingroup$ 1. For Eilenberg-Maclane spaces having the same homotopy groups if enough. $\endgroup$ Sep 6, 2023 at 21:06
  • $\begingroup$ 2. Up to homotopy it suffices to do the realization on non-degenerate simplices (although the visualization is still hard and maybe not helpful?) There $\mathbb Z$ many 1-cells with endpoints glued to the nique 0-cell. Since higher cells are uniquely determined by their spine e.g. $1+5+6+3 = 15$, their attachments are determined by their 1-attachments. For instance the 2-cell $3+6 = 9$ will be attached by going first through the 1-cell of 3, then the 1-cell of 6, then backwards thruogh the 1-cell of 9. $\endgroup$ Sep 6, 2023 at 21:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .