Find the universal covers of these spaces:
\(X=S^1\).
\(Y=S^1\vee S^1\) (two circles glued together along a point).
Solution
The universal cover of \(S^1\) is \[ p:\mathbb R\to S^1, \qquad p(t)=e^{2\pi i t}. \] The space \(\mathbb R\) is simply connected, and every point of \(S^1\) has a small arc whose inverse image is a disjoint union of intervals mapped homeomorphically onto that arc.
The universal cover of \(S^1\vee S^1\) is the \(4\)-regular tree that is the Cayley graph of the free group \[ F(a,b)=\pi_1(S^1\vee S^1). \] Its vertices are group elements \(g\in F(a,b)\), with an \(a\)-edge from \(g\) to \(ga\) and a \(b\)-edge from \(g\) to \(gb\). Map every \(a\)-edge homeomorphically to the first circle and every \(b\)-edge to the second. The resulting graph is a tree, hence simply connected, and the map is a covering map.