Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
6  
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. –  Zhen Lin Sep 17 '13 at 9:18
    
What on earth do you mean by "the standard way "? –  Tim Porter Sep 22 '13 at 7:18

1 Answer 1

up vote 2 down vote accepted

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.

share|improve this answer
    
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. ;) –  Werner Thumann Sep 18 '13 at 9:05
3  
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). –  Peter May Sep 18 '13 at 13:47

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.