Let $G$ be a group having presentation $\langle g_i|r_i\rangle$ with $g_i$ generators and $r_i$ relations. One can construct its associated topological space $X_G$ by wedging $S^1$ and putting $2$-cells to each relation that has corresponding words.

$X_G$ is path connected, semi local simply connected, local path connected. Thus $X_G$ has simply connected universal covering.

$\textbf{Q:}$ Why $X_G$ itself is not Eilenberg Maclane space?(i.e. It is not $K(G,1)$.) Hatcher went through construction of $BG$ space and showed this space is $K(G,1)$.

Ref. Hatcher, Algebraic Topology Chpt 1, Appendix 1.B Example 1B.7

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    $\begingroup$ You might have misread the construction in Hatcher, it does not stop after attaching 2-cells. Once $X_G$ is constructed, one considers $\pi_2(X_G)$, and attaches 3-cells to kill $\pi_2$. One then considers $\pi_3$, and attaches $4$-cells to kill $\pi_3$. And so on inductively. $\endgroup$
    – Lee Mosher
    Commented Jun 15, 2019 at 21:42

1 Answer 1


Consider $G=\mathbb{Z}/2$, it has the presentation $\{g: g^2=1\}$. To define $X_G$ one attaches a $2$-cell along the boundary of $S^1$ for the relation $g^2=1$ the resulting space is the projective plane $\mathbb{R}P^2$ whose universal cover $S^2$ is simply connected, but $\pi_2(S^2)=\mathbb{Z}=\pi_2(\mathbb{R}P^2)$. We deduce that $\mathbb{R}P^2$ is not an Eilenberg McLane space.


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