A sufficient condition for $\pi_1(\operatorname B\Gamma) = 0$

Question

I am currently reading Ghys and Sergiescu's paper Sur un groupe remarquable de difféomorphisms du cercle (French only I'm afraid), but part of their proof of Corollary 3.4 (page 20 of the pdf) is opaque to me.

Some notation first though. Let $K = \operatorname{GA}(\Bbb Q_2)$, the general affine group of the dyadic rationals, and $K_0$ the stabilizer of $0$ with respect to the natural action of $K$ on $\Bbb R$. Let $\Gamma$ denote the pseudogroup on $\Bbb R$ of all local homeomorphisms that are piecewise $K$. Then $\operatorname B\Gamma$ is the classifying space of $\Gamma$.

Ghys and Sergiescu claim that $\pi_1(\mathrm{B}\Gamma) = 0$ and that, to prove this, it suffices to show that the normalizer of $K_0$ is equal to $K$. Why is that sufficient?

Answer 1

Peter Greenberg's paper Pseudogroups from group actions held the key.

There is a weak equivalence between $\mathrm{B}\Gamma$ and $(\mathrm{B}K_0 \ast \mathrm{B}K_0) \vee (\mathrm{B}K/\mathrm{B}K_0)$, where $\ast$ is the join, $\vee$ is the usual wedge sum, and $\mathrm{B}K/\mathrm{B}K_0$ denotes the homotopy pushout of the diagram $$ \mathrm{C}\mathrm{B}K_0 \leftarrow \mathrm{B}K_0 \rightarrow \mathrm{B}K, $$ where $\mathrm{C}$ denotes the cone of a space.

As $K_0 \cong \Bbb Z$ this becomes $\mathrm{B}\Gamma \simeq S^3 \vee (\mathrm{B}K/\mathrm{B}K_0)$, so one need only compute the fundamental group of $\mathrm{B}K/\mathrm{B}K_0$. A quick invocation of Seifert-van Kampen reveals that $$ \pi_1(\mathrm{B}K/\mathrm{B}K_0) \cong \pi_1(\mathrm{B}K)/N(\pi_1(\mathrm{B}K_0)) \cong K/N(K_0). $$