Let \(p : \widetilde{X} \to X\) be a covering map, with \(X\) and \(\widetilde{X}\) locally path-connected, and \(x_0 \in X\). Show that for \(y_0, y_1 \in p^{-1}(\{x_0\})\), there is a deck transformation of \(\widetilde{X}\) taking \(y_0\) to \(y_1\) if and only if \[p_*(\pi_1(\widetilde{X},y_0)) = p_*(\pi_1(\widetilde{X},y_1)).\]
Solution
For \(j=0,1\), let \(C_j\) be the connected component of \(\widetilde X\) containing \(y_j\), and let \(X_0\) be the connected component of \(X\) containing \(x_0\).
1. Each \(C_j\) and \(X_0\) is open and path-connected, and \(p_j\coloneqq p|_{C_j}\colon C_j\to X_0\) is a covering map.
In a locally path-connected space, connected components are open and path-connected. Hence \(C_j\) and \(X_0\) have these properties.
The image of \(C_j\) under \(p\) is all of \(X_0\). Indeed, if \(x\in X_0\), choose a path in \(X_0\) from \(x_0\) to \(x\). Its lift starting at \(y_j\) remains in \(C_j\) and ends over \(x\).
Finally, around any \(x\in X_0\), choose a path-connected evenly covered neighborhood \(U\subseteq X_0\). Every sheet over \(U\) is path-connected, so each sheet that meets \(C_j\) lies entirely in \(C_j\). Thus the sheets lying in \(C_j\) evenly cover \(U\) for \(p_j\).
2. If a deck transformation \(T\colon\widetilde X\to\widetilde X\) satisfies \(T(y_0)=y_1\), then \(p_*(\pi_1(\widetilde X,y_0))=p_*(\pi_1(\widetilde X,y_1))\).
Since \(T\) is a deck transformation, \(p\circ T=p\), so \(p_*\circ T_*=p_*\) with the indicated basepoints. Because \(T\) is a homeomorphism taking \(y_0\) to \(y_1\), \(T_*\colon\pi_1(\widetilde X,y_0)\to\pi_1(\widetilde X,y_1)\) is an isomorphism. Therefore \[\begin{aligned} p_*(\pi_1(\widetilde X,y_0)) &=p_*\bigl(T_*(\pi_1(\widetilde X,y_0))\bigr)\\ &=p_*(\pi_1(\widetilde X,y_1)). \end{aligned}\]
3. If \(p_*(\pi_1(\widetilde X,y_0))=p_*(\pi_1(\widetilde X,y_1))\), then there is a map \(h\colon C_0\to C_1\) with \(p_1\circ h=p_0\) and \(h(y_0)=y_1\).
Every loop in \(\widetilde X\) based at \(y_j\) lies in \(C_j\), so \(\pi_1(\widetilde X,y_j)=\pi_1(C_j,y_j)\). The assumed equality therefore gives \((p_0)_*(\pi_1(C_0,y_0))\subseteq(p_1)_*(\pi_1(C_1,y_1))\). By step 1, \(C_0\) is path-connected and locally path-connected and \(p_1\colon C_1\to X_0\) is a covering map. The lifting criterion for covering spaces therefore gives a lift \(h\) of \(p_0\) through \(p_1\) with \(h(y_0)=y_1\).
4. The map \(h\) of step 3 is a homeomorphism \(C_0\to C_1\) with \(p_1\circ h=p_0\).
The assumed equality also gives the reverse inclusion, so the lifting criterion gives \(k\colon C_1\to C_0\) with \(k(y_1)=y_0\) and \(p_0\circ k=p_1\). Then \(k\circ h\) and \(\operatorname{id}_{C_0}\) are two lifts of \(p_0\colon C_0\to X_0\) through the covering \(p_0\colon C_0\to X_0\), and both take \(y_0\) to \(y_0\). Since \(C_0\) is connected, uniqueness of lifts gives \(k\circ h=\operatorname{id}_{C_0}\). Similarly, \(h\circ k=\operatorname{id}_{C_1}\). Hence \(h\) is a homeomorphism with inverse \(k\).
5. The homeomorphism \(h\) extends to a deck transformation \(T\) of \(\widetilde X\) with \(T(y_0)=y_1\).
Every connected component of \(\widetilde X\) is open, because \(\widetilde X\) is locally path-connected.
If \(C_0=C_1\), define \(T\) to equal \(h\) on \(C_0\) and the identity on every other component.
If \(C_0\neq C_1\), define \(T\) to equal \(h\) on \(C_0\), \(h^{-1}\) on \(C_1\), and the identity on every other component.
Because the components are disjoint open sets, \(T\) is continuous, and so is its inverse, which is defined in the same way with \(h\) and \(h^{-1}\) exchanged. In either case \(p\circ T=p\), since \(p_1\circ h=p_0\) and \(p_0\circ h^{-1}=p_1\) by step 4, and \(T(y_0)=y_1\).
6. Q.E.D.
Step 2 proves that a deck transformation taking \(y_0\) to \(y_1\) forces \(p_*(\pi_1(\widetilde X,y_0))=p_*(\pi_1(\widetilde X,y_1))\). Steps 3, 4, and 5 prove the converse.