Let \(p:\widetilde X\to X\) be a covering space with \(\widetilde X\) path-connected. Let \(A\) be a subspace of \(X\) with inclusion map \(i:A\hookrightarrow X\), and suppose that for \(a\in A\) the induced homomorphism \(i_*:\pi_1(A,a)\to\pi_1(X,a)\) is a surjection. Show that if \(r,s\in\widetilde X\) and \(p(r)=p(s)=a\), then there is a path from \(r\) to \(s\) in the subspace \(p^{-1}(A)\) of \(\widetilde X\).
Solution
1. There is a path\[ \widetilde\alpha:[0,1]\longrightarrow\widetilde X \]from \(r\) to \(s\).
The covering space \(\widetilde X\) is path-connected by hypothesis, so any two of its points can be joined by a path. Apply this to \(r\) and \(s\).
2. The projected path\[ \alpha=p\circ\widetilde\alpha \]is a loop in \(X\) based at \(a\).
Using 1 and the hypotheses on \(r,s\), \[ \alpha(0)=p(r)=a \qquad\text{and}\qquad \alpha(1)=p(s)=a. \] Thus \(\alpha\) begins and ends at \(a\).
3. There is a loop\[ \beta:[0,1]\longrightarrow A \]based at \(a\) whose image in \(\pi_1(X,a)\) is \([\alpha]\).
Surjectivity of \[ i_*:\pi_1(A,a)\longrightarrow\pi_1(X,a) \] implies that the class \([\alpha]\in\pi_1(X,a)\) is \(i_*[\beta]\) for some class \([\beta]\in\pi_1(A,a)\). Choose a loop \(\beta\) in \(A\) representing that class. Viewed as a loop in \(X\) through the inclusion \(i\), it satisfies \[ [i\circ\beta]=[\alpha]. \]
4. There is a homotopy\[ H:[0,1]\times[0,1]\longrightarrow X \]such that\[ H(t,0)=\alpha(t), \qquad H(t,1)=\beta(t), \]and\[ H(0,u)=H(1,u)=a \]for every \(u\in[0,1]\).
By 3, the loops \(\alpha\) and \(\beta\) represent the same element of \(\pi_1(X,a)\). By definition of equality in the fundamental group, they are homotopic relative to the basepoint, which is exactly the stated homotopy.
5. The homotopy \(H\) has a unique lift\[ \widetilde H:[0,1]\times[0,1]\longrightarrow\widetilde X \]with\[ \widetilde H(t,0)=\widetilde\alpha(t). \]
4. The homotopy lifting theorem for covering maps therefore lifts \(H\) uniquely once its lift along the initial edge \([0,1]\times\{0\}\) is prescribed as \(\widetilde\alpha\).
6. For every \(u\in[0,1]\),\[ \widetilde H(0,u)=r \qquad\text{and}\qquad \widetilde H(1,u)=s. \]
By <1>4–<1>5, \[ p(\widetilde H(0,u))=H(0,u)=a, \] so the path \[ u\longmapsto\widetilde H(0,u) \] lies entirely in the fiber \(p^{-1}(a)\). A fiber of a covering map is discrete: if \(U\) is an evenly covered neighborhood of \(a\), each sheet over \(U\) contains at most one point of \(p^{-1}(a)\), so every fiber point is isolated in the fiber subspace. The continuous image of the connected interval \([0,1]\) in a discrete space is a singleton. At \(u=0\) this path equals \[ \widetilde H(0,0)=\widetilde\alpha(0)=r, \] so it is constantly \(r\).
The same argument applied to \[ u\longmapsto\widetilde H(1,u) \] shows it is constant in \(p^{-1}(a)\); at \(u=0\) it equals \[ \widetilde\alpha(1)=s. \] Hence it is constantly \(s\).
7. The path\[ \widetilde\beta(t)=\widetilde H(t,1) \]joins \(r\) to \(s\) and lies in \(p^{-1}(A)\).
By 6, \[ \widetilde\beta(0)=r, \qquad \widetilde\beta(1)=s. \] Moreover <1>4–<1>5 give \[ p(\widetilde\beta(t)) = p(\widetilde H(t,1)) = H(t,1) = \beta(t). \] Since \(\beta(t)\in A\) for every \(t\) by 3, \[ \widetilde\beta(t)\in p^{-1}(A) \] for every \(t\). Thus \(\widetilde\beta\) is the required path in \(p^{-1}(A)\) from \(r\) to \(s\).