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?
$\begingroup$
$\endgroup$
5
-
8$\begingroup$ 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 $G \to \pi$. $\endgroup$– Johannes EbertCommented Dec 7, 2011 at 11:16
-
$\begingroup$ Wouldn't the choice of $x$ affect what $\pi = Aut(x)$ is? $\endgroup$– Gao 2ManCommented Dec 7, 2011 at 13:32
-
3$\begingroup$ Johannes: why do you post this as a comment and not as an answer? $\endgroup$– André HenriquesCommented Dec 7, 2011 at 13:44
-
$\begingroup$ @ 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 $x\to y$. $\endgroup$– André HenriquesCommented Dec 7, 2011 at 13:45
-
4$\begingroup$ 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.) $\endgroup$– Tim PorterCommented Dec 7, 2011 at 15:53
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
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.
answered Jan 23, 2012 at 20:45