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Let $\mathcal{C}$ be a groupoid, and let $B \mathcal{C}:= |N\mathcal{C}|$ be its classifying space, i.e. the geometric realisation of its nerve.

How to prove that $B \mathcal{C}$ is a disjoint union of $K(\pi,1)$'s? Ultimately I would like to give a precise argument to show that the classifying space construction induces an equivalence $\mathrm{Ho}(\mathsf{Grpd}) \simeq \mathrm{Ho}(\mathsf{Top}^{CW}_{\leq 1})$, see e.g. this ncatLab entry.

Minkowski
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    You've not specified what $\pi$ is, but also this statement is wrong. A groupoid $\mathcal{C}$ is a disjoint union of its components, each of which is equivalent to $BG$ for a group $G$. The geometric realization of $BG$ is a $K(G,1)$ and the geometric realization of $B\mathcal{C}$ will be the likewise disjoint union of these. As for why $BG$ is a $K(G,1)$, this has been explained many times, see e.g. this question. – Thorgott Feb 11 '25 at 13:47
  • You're absolutely right, it was easier than what I thought, and indeed my claim was wrong and I edited my question accordingly, but you already answered it. Thanks, that's clear now! – Minkowski Feb 11 '25 at 14:13

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