(Purdue Aug ’09) Let \(p:E\to B\) be a covering map. Let \(Y\) be locally path-connected. Let \(g:Y\to E\) be a function such that
\(p\circ g\) is continuous;
\(g\circ\gamma\) is continuous for every path \(\gamma\) in \(Y\).
Prove that \(g\) is continuous.
Solution
Let \(h=p\circ g:Y\to B\), which is continuous by hypothesis. Fix \(y_0\in Y\), and choose an evenly covered neighborhood \(U\) of \(h(y_0)\). Let \(V_0\) be the sheet over \(U\) containing \(g(y_0)\), and let \[ s:U\longrightarrow V_0 \] be the inverse of \(p|_{V_0}\).
By continuity of \(h\), \(h^{-1}(U)\) is a neighborhood of \(y_0\). Since \(Y\) is locally path connected, choose a path-connected open neighborhood \[ y_0\in W\subset h^{-1}(U). \] We claim that on \(W\), \[ g=s\circ h. \] Take \(y\in W\), and choose a path \(\gamma:[0,1]\to W\) from \(y_0\) to \(y\). By hypothesis \(g\circ\gamma\) is continuous. Both \[ g\circ\gamma \quad\text{and}\quad s\circ h\circ\gamma \] are lifts of the same path \(h\circ\gamma\), and they agree at \(0\), since both take the value \(g(y_0)\). Uniqueness of path lifting gives equality on all of \([0,1]\), in particular \[ g(y)=s(h(y)). \] Thus \(g|_W=s\circ h|_W\), which is continuous. Since \(y_0\) was arbitrary, \(g\) is continuous on \(Y\).