Visualizing $|\mathcal{B}\mathbb{Z}| \simeq S^1$.

Question

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 $f:X\to Y, g:Y\to X$.

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 $K(1,\mathbb{Z})$.

Proof:

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.