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It is very well know that groupoids, considered as spaces via the nerve construction, are homotopy 1-types, i.e. aspherical. Here is a sketch of proof: Consider the canonical functor $f:C\rightarrow \pi_1(C)$ for a category $C$ and its fundamental groupoid $\pi_1(C)$. The Quillen-fiber based at some object $Y$, i.e. the slice category $Y\backslash f$, is the universal covering category of $C$. If $C=G$ is a groupoid, then $\pi_1(G)$ is $G$ itself and $f$ is essentially the identity. It is not hard to see that the Quillen-fiber in this case has a terminal object and thus is contractible, which proves the statement.

Is this the standard way to prove that groupoids have vanishing higher homotopy groups?

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    $\begingroup$ How about this? It is well-known that the nerve of a groupoid is a Kan complex, and that the nerve of any category is 2-coskeletal. But 2-coskeletal Kan complexes must have trivial higher homotopy groups. $\endgroup$ – Zhen Lin Sep 17 '13 at 9:18
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    $\begingroup$ What on earth do you mean by "the standard way "? $\endgroup$ – Tim Porter Sep 22 '13 at 7:18
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A low tech-argument:

-a groupoid is a disjoint union of connected groupoids

-the nerve construction preserves disjoint unions

-a connected groupoid is equivalent to a group, and thus its nerve is aspherical.

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  • $\begingroup$ Ah well, I know that a group being aspherical is even more standard than a groupoid being aspherical. But in the spirit of the question you should also give a nice argument for this case. ;) $\endgroup$ – Werner Thumann Sep 18 '13 at 9:05
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    $\begingroup$ A good construction of the classifying space BG of a group comes with a construction of a universal cover EG >--> BG with the discrete group G as fiber. Therefore BG = K(G,1). $\endgroup$ – Peter May Sep 18 '13 at 13:47

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