Let $G$ be a discrete group. $EG$ is defined as the $\triangle$-complex (Hatcher p.102) whose $n$-simplices are given by $[g_0,g_1,...,g_n]$ glued together in the obvious way. Then define $BG=EG/G$. I am trying to understand this definition by a simple example. If I choose $G= \mathbb{Z}/2\mathbb{Z}$, then I think $EG$ should be a path with endpoints $0$ and $1$, and then $BG$ will identify the endpoints and give us $S^1$. This does not seem right since $S^1 = K(\mathbb{Z},1)$. What am I doing wrong here?


1 Answer 1


In your construction, $E(\mathbb{Z}/2\mathbb{Z})$ is $S^\infty$ with the antipodal $\mathbb{Z}/2\mathbb{Z}$-action.

You can think of $n$-simplices here as binary strings of length $n+1$. Nondegenerate simplices are given by binary strings that alternate strictly between $0$ and $1$.

  • There are two $0$-simplices $[0]$ and $[1]$. So far this is $S^0$.
  • There are two (nondegenerate) $1$-simplices $[01]$ and $[10]$ which are attached to their boundaries: $[0]$ at one end, $[1]$ at the other. This is $S^1$.
  • There are two (nondegenerate) $2$-simplices $[010]$ and $[101]$, which are attached to $[01]$ and $[10]$ (and other degenerate simplices). This is $S^2$.

This pattern continues, and you may recognize the whole thing as the usual presentation of $S^\infty$ as a $\mathbb{Z}/2\mathbb{Z}$-CW complex.

The quotient of $S^\infty$ by the antipodal action is $S^\infty / (\mathbb{Z}/2\mathbb{Z}) \simeq \mathbb{R}P^\infty$, which is indeed $K(\mathbb{Z}/2\mathbb{Z}, 1)$.

  • $\begingroup$ I still have 1 question: you say nondegenerate simplices are given by binary strings alternating between 0 and 1. Why is that? [0,1,0] has 3 faces, [0,1], [1,0] and [0,0], but so does [0,0,1]. $\endgroup$ Nov 20, 2020 at 2:42
  • 1
    $\begingroup$ The $2$-simplex $[0,0,1]$ only has faces $[0,0]$ and $[0,1]$ (twice), so visually it's "collapsed" onto the edge $[0,1]$. $\endgroup$
    – JHF
    Nov 20, 2020 at 2:48

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