Show that for a covering space action of a group \(G\) on a simply-connected space \(Y\), \(\pi_1(Y/G)\) is isomorphic to \(G\). [If \(Y\) is locally path-connected, this is a special case of part (c) of Proposition 1.40.]
Solution
Let \[ q:Y\to Y/G \] be the orbit map.
1. The map \(q\) is a covering map with deck transformation group \(G\).
This is exactly the definition and basic property of a covering space action: the translates of a suitable neighborhood are pairwise disjoint and map homeomorphically to one neighborhood in the quotient. Since \(Y\) is path connected, every deck transformation of \(q\) is determined by the image of one point, and Proposition 1.40 gives \[ \operatorname{Deck}(Y/(Y/G))\cong G. \]
2. Since \(Y\) is simply connected, \(q\) is the universal covering map of \(Y/G\).
A simply connected covering space is universal.
3. Therefore\[ \boxed{\pi_1(Y/G)\cong G.} \]
For a universal cover, the deck transformation group is canonically isomorphic to the fundamental group of the base, after choosing a basepoint in the fiber. Combining this with 1 gives the displayed isomorphism.