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

Let $G$ be a connected groupoid. Is the nerve $BG$ a $K(\pi, 1)$, and if so, is there a groupoid homomorphism $f:G\to \pi$ that induces the homotopy equivalence?

share|improve this question
8  
If $x$ is an object of $G$, then $\pi:=Aut(x) \to G$ is an equivalence of categories and so induces a homotopy equivalence $B \pi \to BG$. For the inverse, choose for each object $y$ of $G$ amorphism $c_xx:x \to y$. Assigning to a morphism $d:y \to z$ the endomorphism $c_{z}^{-1} d c_x$ of $ x$ is a functor $G \to \pi$. –  Johannes Ebert Dec 7 '11 at 11:16
    
Wouldn't the choice of $x$ affect what $\pi = Aut(x)$ is? –  Gao 2Man Dec 7 '11 at 13:32
3  
Johannes: why do you post this as a comment and not as an answer? –  André Henriques Dec 7 '11 at 13:44
    
@ Gao 2Man: In a connected groupois, all $Aut(x)$ are isomorphic. The isomorphism $Aut(x)\to Aut(y)$ is given by conjugation by an arrow $x\to y$. –  André Henriques Dec 7 '11 at 13:45
4  
Adding to what André says. The choice of $x$ does not affect the isomorphism type of $Aut(x)$ BUT different choices of the connecting morphisms $c_x$ in Johannes comment will give different isomorphisms. (This is why the groupoid situation can sometimes be more fruitful to study than merely the group one, as the automorphism groups of different objects are more in evidence.) –  Tim Porter Dec 7 '11 at 15:53
add comment

1 Answer 1

Another way of putting this is that there is a notion of $K(G,1)$ for $G$ a groupoid, in fact this is just the classifying space $BG$ of the groupoid, i.e. the realisation of the simplicial nerve of $G$. As pointed out by Segal, the nerve construction on categories (or groupoids) takes equivalences to homotopy equivalences.

share|improve this answer
add comment

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.