Proof. We will prove existence and uniqueness.

Existence. Since \(p:\widetilde{X}\to X\) is a covering space, each \(f(t)\in X\) has an evenly covered neighborhood \(U_t\). Then, \[\{f^{-1}(U_t) \operatorname{\big|}t\in I\}\] is an open cover of \(I\), so by the Lebesgue number lemma there exists \(\varepsilon>0\) such that each \(\varepsilon\)-ball \(B_\varepsilon(s)\subseteq I\) is contained in some \(f^{-1}(U_t)\). We choose \(n\in \mathbb{N}\) such that \(1/n < \varepsilon\). Then, we have \[I = \left[0,\frac{1}{n}\right] \cup \left[\frac{1}{n},\frac{2}{n}\right] \cup \cdots \cup \left[\frac{n-1}{n},\frac{n}{n}\right]\] where each piece has \[f\left(\left[\frac{k-1}{n},\frac{k}{n}\right]\right) \subseteq U_t\] for some \(t\in I\), and we denote this \(U_t\) by \(U_k\).

Now, we use these pieces to lift \(f\). We inductively define \(\widetilde{f_k}: [0,k/n], 0 \to \widetilde{X},\widetilde{x}\) with \(p\circ \widetilde{f_k} = f|_{[0,k/n]}\). Our base case requires us to define \(\widetilde{f_0}: [0,0], 0 \to \widetilde{X}, \widetilde{x}\), and we can just set \(\widetilde{f_0}(0)=\widetilde{x}\).

Now, assuming we’ve defined \(\widetilde{f_{k-1}}\), we define \(\widetilde{f_k}\). We choose the connected component \(\widetilde{U_k}\) of \(p^{-1}(U_k)\) that satisfies \[\widetilde{f_{k-1}}(\frac{k-1}{n}) \in \widetilde{U_k},\] and we then define \[\widetilde{f_{k}}(t) = \begin{cases} \widetilde{f_{k-1}}(t) & t\in \left[0,\frac{k-1}{n}\right]\\ \left(p|_{\widetilde{U_k}}\right)^{-1}(f(t)) & t\in \left[\frac{k-1}{n},\frac{k}{n}\right]. \end{cases}\] This map \(\widetilde{f_k}\) is well-defined because \(p\) restricted to \(\widetilde{U_k}\) is a homeomorphism and has an inverse (since \(U_k\) is chosen to be evenly covered), and by choice of \(\widetilde{U_k}\) and definition of \(\widetilde{f_{k-1}}\) we have \[\left(p|_{\widetilde{U_k}}\right)\left(\widetilde{f_{k-1}}(\frac{k-1}{n})\right) = f(\frac{k-1}{n}),\] so we can apply \((p|_{\widetilde{U_k}})^{-1}\) to get that both parts of \(\widetilde{f_{k}}\) agree on the overlap.

Since both parts are continuous and agree on overlap, by the pasting lemma (for closed sets), \(\widetilde{f_k}\) is continuous. Furthermore, it satisfies \(p\circ \widetilde{f_k} = f|_{[0,k/n]}\) because we can check on both parts of \(\widetilde{f_k}\).

Once we get to the \(n\)th step, we get our desired lift \(\widetilde{f}: I,0\to \widetilde{X},\widetilde{x}\).

Uniqueness. Suppose \(\widetilde{g}:I,0\to \widetilde{X},\widetilde{x}\) is another lift of \(f\). Our strategy to show \(\widetilde{f}=\widetilde{g}\) is to show that each time we extended \(\widetilde{f_k}\) that our hand was forced—we didn’t have any other options.

We prove that \[\widetilde{f}|_{[0,k/n]} = \widetilde{g}|_{[0,k/n]}\] by induction on \(k\). For our base case, by assumption we have \(\widetilde{f}(0)=\widetilde{g}(0)\). Now, assume that we know for \(k\) that \[\widetilde{f}|_{[0,(k-1)/n]} = \widetilde{g}|_{[0,(k-1)/n]}.\] We know that \[p \circ \widetilde{g} = f,\] and on \([(k-1)/n,k/n]\) our map \(f\) lands inside \(U_k\). Therefore, on \([(k-1)/n,k/n]\) \(\widetilde{g}\) lands in \(p^{-1}(U_k)\). Since \(\widetilde{g}\) is continuous, the image the connected set \([(k-1)/n,k/n]\) under \(\widetilde{g}\) must be connected, i.e., it must land in a single connected component of \(p^{-1}(U_k)\).

By assumption, \(\widetilde{f}((k-1)/n)=\widetilde{g}((k-1)/n)\), so the image of \([(k-1)/n,k/n]\) under \(\widetilde{g}\) must be \(\widetilde{U_k}\), the same connected component as \(\widetilde{f}\). Therefore, we can write \[p|_{\widetilde{U_k}} \circ \widetilde{f}|_{[(k-1)/n,k/n]} = p|_{\widetilde{U_k}} \circ \widetilde{g}|_{[(k-1)/n,k/n]},\] and since \(p|_{\widetilde{U_k}}\) a homeomorphism, we have \[\widetilde{f}|_{[(k-1)/n,k/n]} = \widetilde{g}|_{[(k-1)/n,k/n]}.\] Thus, we have completed our induction step, having shown \[\widetilde{f}|_{[0,k/n]} = \widetilde{g}|_{[0,k/n]}.\] After \(n\) steps, we get \(\widetilde{f}=\widetilde{g}\). Thus, the lift is unique. ◻

Proof. Let \(a,b\in X\), and let \(f: I\to X\) be a path \(a\) to \(b\).

We attempt to define a bijection \(\varphi: p^{-1}(a) \to p^{-1}(b)\). For each \(\widetilde{a_i}\in p^{-1}(x)\), by the path lifting property, there exists a unique lift \[\widetilde{f_i}: I,0 \to \widetilde{X},\widetilde{a_i}\] of \(f\). Set \(\varphi(a)=\widetilde{f_i}(1)\), i.e., the other endpoint of the lifted path.

The function \(\varphi\) is well-defined by the uniqueness of the lifted path. We need to show that it is injective/surjective. Let \(\widetilde{a_1},\widetilde{a_2} \in p^{-1}(a)\), and assume that \(\varphi(\widetilde{a_1})=\varphi(\widetilde{a_2})\), so \(\widetilde{f_1}(1)=\widetilde{f_2}(1)\). Then, the reversed lifts \[\overline{\widetilde{f_1}}, \overline{\widetilde{f_2}}\] are both lifts of the reversed path \(\widetilde{f}\). Since the reversed lifts both have the same starting point, by the uniqueness part of the path-lifting property, we must have \[\overline{\widetilde{f_1}} = \overline{\widetilde{f_2}},\] so \(\widetilde{f_1}=\widetilde{f_2}\), so \(\widetilde{a_1}=\widetilde{a_2}\). Thus, \(\varphi\) is injective.

Now, we check surjectivity. Let \(\widetilde{b_j}\in p^{-1}(b)\). Then, we can lift the reversed path \(\overline{f}\) starting at \(\widetilde{b_j}\) to the lift \(\widetilde{\overline{f}_j}\), and we can check that \(\widetilde{\overline{f}_j}(1) \in p^{-1}(a)\) maps to \(\widetilde{b_j}\) under \(\varphi\). ◻

Proof. Let \(y\in Y\) be given. For each \(t\in I\), there exists an evenly covered neighborhood \(U_t\) of \(F(y,t)\). Then, there exists a basic neighborhood \(N_t\times V_t \subseteq F^{-1}(U_t)\) of \((y,t)\). The collection \[\{V_t \operatorname{\big|}t\in I\}\] is a cover of \(I\) which has subcover \(V_{t_1},\dots,V_{t_m}\). Choose a Lebesgue number \(\varepsilon>0\) for the subcover \(\{V_{t_i}\}\) and choose \(n\in \mathbb{N}\) so that \(1/n<\varepsilon\). Thus, each \[I_k = \left[\frac{k-1}{n}, \frac{k}{n}\right]\] is contained in some \(V_{t_i}\). Then, we can take \[N = \bigcap_{i=1}^m N_{t_i},\] so that for each \(k\) we have \[N\times I_k \subseteq F^{-1}(U_{t_k}).\] This is a big win—we’ve have a neighborhood \(N\) of \(y\) such that \(N\times I\) can be written as the union of sets \(N\times I_k\) which map into evenly covered sets.

Now, we’re in a good position to lift \(F|_{N\times I}\) (this is an abuse of notation—we don’t really mean the restriction because for different \(N\) this map a-priori may not be the same. Eventually, we will prove that these lifts all agree with each other, and the abuse of notation will be correct). Inductively, we define \[\widetilde{F}|_{N\times [0,k/n]},\] with our base case \(\widetilde{F}|_{N\times [0,0]}\) given by assumption, and our induction step by choosing a connected component \(\widetilde{V_{t_i}}\subseteq p^{-1}(V_{t_i})\) containing \(F(y,(k-1)/n)\) and defining \[\widetilde{F}|_{N\times [(k-1)/n,k/n]} = \left(p|_{\widetilde{V_{t_i}}}\right)^{-1}\circ F|_{[(k-1)/n,k/n]}.\] The resulting map will be well-defined by choice of \(\widetilde{V_{t_i}}\) and continuous by the pasting lemma (for closed sets).

Now, for each \(y\in Y\) we have neighborhoods \(N_y=N\) and a lift \[\widetilde{F}|_{N_y\times I} : N_y\times I \to \widetilde{X}\] with \(\widetilde{F}_{N_y\times \{0\}}\) given by assumption. We need to show that these lifts agree with each other so that they assemble into a continuous map \(\widetilde{F}: Y\times I\to \widetilde{X}\).

The collection \[\{N_y \operatorname{\big|}y\in Y\}\] forms an open cover of \(Y\). Let \(y,y'\in Y\) and let \(z\in N_y\cap N_{y'}\). Then, \[\left(\widetilde{F}_{N_y\times I}\right)_{\{z\}\times I}: \{z\}\times I\to \widetilde{X}, \qquad \left(\widetilde{F}_{N_{y'}\times I}\right)_{\{z\}\times I}: \{z\}\times I\to \widetilde{X}\] are both lifts of the path \[F|_{\{z\}\times I}: \{z\}\times I \to X\] that send \(0\mapsto \widetilde{F}_0(z)\). Thus, by the uniqueness of the path lifting property, these two maps must be the same. This just tells us that any two lifts agree on intersection, so we can get a well-defined lift \(\widetilde{F}: Y\times I \to \widetilde{X}\), which is continuous by the pasting lemma (for open sets).

Finally, our lift \(\widetilde{F}: Y\times I\to \widetilde{X}\) is unique again because each restriction to \(\{y\}\times I\) is a lift of the path \(F|_{\{y\}\times I}\), so unique by the path lifting property. ◻

Proof. By the path lifting property, \(F_0:I\to X\) lifts to a unique path \(\widetilde{F}_0: I\to \widetilde{X}\) starting at \(\widetilde{x}\). Then, by the homotopy lifting property, \(F\) lifts to a unique map \[\widetilde{F}: I\times I \to \widetilde{X}\] with the already defined \(\widetilde{F}_0\).

We now show that \(\widetilde{F}\) is a homotopy of paths. Since the paths \[t \mapsto F_t(0), \qquad t\mapsto F_t(1)\] are constant (because \(F\) is a homotopy of paths), and we have that \[t\mapsto \widetilde{F}_0(0), \qquad t\mapsto \widetilde{F}_0(1)\] are lifts of the respective paths, by uniqueness of path lifting we must have \(\widetilde{F}_t(0)\) and \(\widetilde{F}_t(1)\) constant. Thus, \(\widetilde{F}\) is a homotopy of paths. ◻

Proof. Let \([\widetilde{f}]\in \ker{p_*}\), so \(p\circ \widetilde{f} \simeq c_x\) rel \(\partial I\). Let \[F: I\times I \to X, \qquad F_0 = p\circ \widetilde{f}, \qquad F_1=c_x\] be the homotopy. We can then lift this homotopy of paths to a homotopy of paths based at \(\widetilde{x}\), giving us \[\widetilde{F}: I\times I \to X, \qquad \widetilde{F}_0 = \widetilde{f}.\] We claim \(\widetilde{F}_1 = c_{\widetilde{x}}\). This must be the case because \(\widetilde{F}_1\) is a lift of \(c_x\) based at \(\widetilde{x}\), and so is \(c_{\widetilde{x}}\). Thus, by uniqueness \(\widetilde{F}_1 = c_{\widetilde{x}}\).

Therefore, \([\widetilde{f}]=0\) because \(\widetilde{f}\) is homotopic to \(c_{\widetilde{x}}\). Thus, \(p_*\) is injective. ◻

Proof. Let \(G=\pi_1(X,x)\) and \(H=p_*(\pi_1(\widetilde{X},\widetilde{x}))\). Our strategy is to define a bijection from \([G:H]\) to \(p^{-1}(x)\), since \(\deg(p)=|p^{-1}(x)|\). We define \[\begin{aligned} \theta: \pi_1(X,x)/H &\longrightarrow p^{-1}(x) \end{aligned}\] as follows. Given \([\alpha]H\), first lift \(\alpha\) to get \(\widetilde{\alpha}\) starting at \(\widetilde{x}\). Then, set \(\theta([\alpha])=\widetilde{\alpha}(1)\), i.e., the other endpoint of \(\widetilde{\alpha}\).

We need to show \(\theta\) is well-defined, surjective, and injective. To show well-defined, assume \([\alpha]=[\beta]\), so there is path homotopy \(F:I\times I \to X\). We can lift this homotopy to a path homotopy \(\widetilde{F}\) starting at \(\widetilde{\alpha}\). Then, \(\widetilde{F}_t(1)\) is constantly \(\widetilde{F}_0(1)=\widetilde{\alpha}(1)\) because the lift is a path-homotopy. Thus, we can get \(\theta([\beta])=\widetilde{\alpha}(1)\), and \(\theta\) is well-defined.

Next, we show \(\theta\) is surjective. Let \(\widetilde{y}\in p^{-1}(x)\). Since \(\widetilde{X}\) is path-connected, there is a path \(\widetilde{\alpha}\) from \(\widetilde{x}\) to \(\widetilde{y}\), which projects to \(p\circ \widetilde{\alpha}\), which is a loop at \(x\). Then, by the uniqueness in the path lifting property, we can get \[\theta([p\circ \widetilde{\alpha}])=\widetilde{\alpha}(1)=\widetilde{y}.\]

Finally, we show injectivity. Assume \(\theta([\alpha])=\theta([\beta])\). Then, \[\widetilde{\alpha} \cdot \overline{\widetilde{\beta}}\] is a loop at \(\widetilde{x}\), and \[[\alpha]\cdot [\beta]^{-1} = p(\widetilde{\alpha} \cdot \overline{\widetilde{\beta}}) \in H,\] so \([\alpha]H=[\beta]H\). ◻

Proof. First, assume that a lift \(\widetilde{f}\) exists. Then, \[\begin{aligned} f &= p \circ \widetilde{f}\\ f_* &= p_* \circ \widetilde{f}_*, \end{aligned}\] so \(\mathop{\mathrm{im}}(f_*) \subseteq \mathop{\mathrm{im}}(p_*)\).

Conversely, assume that \(\mathop{\mathrm{im}}(f_*)\subseteq \mathop{\mathrm{im}}(p_*)\). We need to come up with a definition of \(\widetilde{f}\), show that it is well-defined, and finally show that it is continuous. Let \(y'\in Y\), and let \(\gamma\) be a path from \(y\) to \(y'\). Then, \(f\circ \gamma\) is a path from \(f(y)\) to \(f(y')\) in \(X\). By the path lifting property, there exists a lift of \(f\circ \gamma\), which we denote \(\widetilde{f\circ \gamma}\). Then, we set \[\widetilde{f}(y) = (\widetilde{f\circ \gamma})(1).\] A priori, this depends on \(\gamma\). To show that this is actually a function, we need to show it is well-defined.

Let \(\gamma'\) be another path from \(y\) to \(y'\). We will show that \((\widetilde{f\circ \gamma})(1) = (\widetilde{f\circ \gamma'})(1)\). Consider the loop \[(f\circ \gamma') \cdot (f\circ \overline{\gamma}) = f\circ (\gamma' \cdot \overline{\gamma}),\] whose equivalence class is in \(f_*(\pi_1(Y,y)) \subseteq p_*(\pi_1(\widetilde{X},\widetilde{x}))\). Thus, \[[f\circ (\gamma' \cdot \overline{\gamma})] = [p \circ \alpha]\] for some loop \([\alpha] \in \pi_1(\widetilde{X},\widetilde{x})\). Thus, we have a homotopy \(H: I\times I \to X\) with \(H_0 = p\circ \alpha\) and \(H_1 = f\circ (\gamma' \cdot \overline{\gamma})\). Since we have a lift \(\widetilde{H_0} = \alpha\), by the homotopy lifting property we have a lift of the homotopy \(\widetilde{H}\). Then, \(\widetilde{H}_1\) is a lift of \(f\circ (\gamma'\cdot \overline{\gamma})\). In particular, the two halves of \(\widetilde{H}_1\) are (reparameterized) lifts of \(f\circ \gamma'\) and \(f\circ \overline{\gamma}\). By the path lifting property, we then must have \[\widetilde{H}_1 = \left(\overline{f\circ \gamma'}\right) \cdot \left(\overline{\widetilde{f\circ \gamma}}\right),\] so therefore we indeed have \[(\widetilde{f\circ \gamma})(1) = (\widetilde{f\circ \gamma'})(1),\] so \(\widetilde{f}\) is well-defined.

Finally, we show that \(\widetilde{f}\) is continuous. Let \(y'\in Y\). Let \(\widetilde{U}\) be a neighborhood of \(\widetilde{f}(y')\). We can assume \(\widetilde{U}\) is small enough such that \(p|_{\widetilde{U}}\) is a homeomorphism to an open set \(U\subseteq X\).

By local path connectedness, choose an open path connected neighborhood \(V\) of \(y\) such that \(V\subseteq f^{-1}(U)\). We claim that \(\widetilde{f}(V)\subseteq \widetilde{U}\), so that \(\widetilde{f}\) is continuous at \(y'\).

Let \(y''\in V\). To evaluate \(\widetilde{f}\), let \(\gamma:I\to Y\) be a path \(y\) to \(y'\), and let \(\eta:I\to V\) be a path \(y'\) to \(y''\). Then, using the definition of \(\widetilde{f}\) and the uniqueness of path-lifting, \[\widetilde{f}(y'') = \widetilde{f\circ (\gamma\cdot \eta)} (1) = (\widetilde{f\circ \gamma}) \cdot (\widetilde{f\circ \eta})(1) = (\widetilde{f\circ \eta})(1).\] Since \(\eta\) is in \(V\subseteq f^{-1}(U)\), we actually know that the lift will be \[(p|_{\widetilde{U}})^{-1} \circ f \circ \eta\] by the uniqueness of path lifting. Thus, \[\widetilde{f}(y'') = (\widetilde{f\circ \eta})(1) = (p|_{\widetilde{U}})^{-1} \circ f \circ \eta(1) \quad \in \widetilde{U},\] so \(\widetilde{f}\) is continuous. ◻

Proof. Assume \(\widetilde{f_1}(y) = \widetilde{f_2}(y)\) for some point \(y\in Y\). We show that the set of points where \(\widetilde{f_1}\) and \(\widetilde{f_2}\) agree is clopen, so since it is nonempty, it must all of \(Y\) by connectedness.

Let \(y'\in Y\). Our goal is to find a neighborhood of \(y'\) where the two lifts entirely agree or disagree. Let \(U\subseteq X\) be an evenly covered neighborhood of \(f(y')\), and let \(\widetilde{U_1},\widetilde{U_2}\subseteq \widetilde{X}\) be the sheets above \(U\) containing \(\widetilde{f_1}(y'),\widetilde{f_2}(y')\) respectively. By continuity of the lifts, we can choose a neighborhood \(N\subseteq Y\) of \(y'\) such that \[\widetilde{f_1}(N)\subseteq \widetilde{U_1}, \qquad \widetilde{f_2}(N)\subseteq \widetilde{U_2}\] If \(\widetilde{f_1}(y')=\widetilde{f_2}(y')\), then \(\widetilde{U_1}=\widetilde{U_2}=\widetilde{U}\), and \[p|_{\widetilde{U}} \circ \widetilde{f_1}|_{N} = p|_{\widetilde{U}} \circ \widetilde{f_2}|_{N}\] implies by injectivity of \(p|_{\widetilde{U}}\) that \(\widetilde{f_1},\widetilde{f_2}\) agree on all of \(N\). Thus, the set of points where the lifts agree is open.

Alternatively, if \(\widetilde{f_1}(y')\neq \widetilde{f_2}(y')\), then \(\widetilde{U_1}\neq \widetilde{U_2}\), and even more they are disjoint (because they are sheets), so \(\widetilde{f_1},\widetilde{f_2}\) disagree on \(N\). Thus, the set of points where the lifts disagree is open. ◻

Proof. Let \(x\in X\). Then, \(x\) has a neighborhood \(U\) with lift \(\widetilde{U}\) such that \(p|_{\widetilde{U}}\) is a homeomorphism.

Let \(\gamma:I \to U\) be a loop at \(x\). Then, \(\gamma\) lifts to the loop \(p|_{\widetilde{U}}^{-1} \circ \gamma\), which is nullhomotopic in \(\widetilde{X}\) because \(\widetilde{X}\) is simply connected. Thus, there exists a homotopy \(h_t:I\to \widetilde{X}\) with \[h_0 = p|_{\widetilde{U}}^{-1} \circ \gamma\] and \(h_1\) being the constant loop. Then, \(p\circ h_t: I\to X\) is a homotopy in \(X\) which makes \(\gamma\) nullhomotopic. ◻

Proof. Choose a basepoint \(x\in X\). Define the set \[\widetilde{X} = \{[\gamma] \operatorname{\big|}\gamma \text{ a path in } X \text{ starting at } x\},\] where the equivalence class \([\gamma]\) is with respect to path homotopies (fixing endpoints). We then define the map \[\begin{aligned} p: \widetilde{X} &\longmapsto X\\ [\gamma] &\longmapsto \gamma(1). \end{aligned}\] We now need to put a topology on \(\widetilde{X}\), prove that \(p\) is a covering map, and prove that \(\widetilde{X}\) is simply connected.

First, let’s give \(\widetilde{X}\) a topology. We extract data from \(X\). Define \(\mathcal{U}\) to be the collection of path-connected open sets \(U\subseteq X\) such that the induced inclusion map \(\pi_1(U) \to \pi_1(X)\) is trivial (this doesn’t depend on basepoint because \(U\) is path-connected). By the locally path connected and SLSC hypotheses, \(\mathcal{U}\) forms a basis for \(X\).

Now, for each \(U\in \mathcal{U}\) define \[U_{[\gamma]} = \{[\gamma\circ \eta] \operatorname{\big|}\eta \text{ is a path in $U$ with $\eta(0)=\gamma(1)$}\}.\] One can check that the collection of \(U_{[\gamma]}\) is a basis for topology for \(\widetilde{X}\) (i.e., can generate intersections of basic sets and the entire space). One can next check that \(p: U_{[\gamma]} \to U\) is a homeomorphism, which shows that \(p\) is not only continuous but a covering map.

Finally, we show that \(\widetilde{X}\) is simply connected. First, we show path connectedness. Let \([\gamma]\in \widetilde{X}\). Define \(\gamma_t\) by \[s\mapsto \begin{cases} \gamma(s) & s\in [0,t]\\ \gamma(t) & s\in [t,1], \end{cases}\] i.e., \(\gamma_t\) follows \(\gamma\) until \(t\), and then is stationary. Then, we have a path from the equivalence class of the constant loop \([x]\in \widetilde{X}\) to \([\gamma]\) by \[t\mapsto [\gamma_t].\] Then, we show \(\pi_1(\widetilde{X},[x]) = 0\). Let \(\widetilde{\gamma}\) be a loop in \(\widetilde{X}\) at \([x]\), so it is a lift of the path \(\gamma=p\circ \widetilde{\gamma}\). However, \(t\mapsto [\gamma_t]\) also lifts \(\gamma\) starting at \([x]\). By the uniqueness of the path-lifting property, these two paths are equal. In particular, we must have \([\gamma_1]=[x]\) for \(t\to [\gamma_t]\) to be a loop, so \([\gamma] = [x]\). Therefore, \(\pi_1(\widetilde{X},[x])=0\). ◻

Proof Sketch.. Take \(X_H\) to be a quotient of the universal covering space \(\widetilde{X}/\sim\) defined in the proof of theorem 21, where \(\sim\) is the equivalence relation defined by \([\gamma]\sim [\gamma']\) if \(\gamma' \cdot \overline{\gamma} \in H\). ◻

Proof. Assume that \(p_1,p_2\) are isomorphic. Then, we have a homeomorphism \(f: X,x \to X,x\) such that \[p_1 \circ f = p_2,\] so the induced map \(f_*\) will be an isomorphism of \(\mathop{\mathrm{im}}(p_1)_*\) and \(\mathop{\mathrm{im}}(p_2)_*\).

Next, assume that \[{p_1}_*(\pi_1(\widetilde{X}_1,\widetilde{x_1})) = {p_2}_*(\pi_1(\widetilde{X}_2,\widetilde{x_2})).\] By the lifting criterion theorem 17, there exist a unique lifts \[\begin{aligned} \widetilde{p_1}: \widetilde{X_1}, \widetilde{x_1} &\longrightarrow \widetilde{X_2}, \widetilde{x_2}\\ \widetilde{p_2}: \widetilde{X_2}, \widetilde{x_2} &\longrightarrow \widetilde{X_1}, \widetilde{x_1}. \end{aligned}\] We claim that \(\widetilde{p_1},\widetilde{p_2}\) are inverse to each other. We look at the composition \[\widetilde{p_2} \circ \widetilde{p_1}: \widetilde{X_1},\widetilde{x}_1 \longrightarrow \widetilde{X_1},\widetilde{x}_1.\] This is a lift of \(p_1\) to \(\widetilde{X_1}\) fixing \(\widetilde{x}_1\). The identity map \(\mathop{\mathrm{id}}_{\widetilde{X_1}}\) is also a lift, so by uniqueness in proposition 18, we must have \[\widetilde{p_2} \circ \widetilde{p_1} = \mathop{\mathrm{id}}_{\widetilde{X_1}}.\] Similarly, \(\widetilde{p_1} \circ \widetilde{p_2} = \mathop{\mathrm{id}}_{\widetilde{X_2}}\), and these maps \(\widetilde{p_1},\widetilde{p_2}\) are morphisms between covering spaces \(p_1,p_2\) by construction.

Thus, \(p_1,p_2\) are isomorphic. ◻

Proof. We only need to show one inclusion of the equality, since the other inclusion comes from reversing \(\widetilde{x},\widetilde{x}'\). Let \(\alpha\in \pi_1(\widetilde{X},\widetilde{x})\). Then, \[\overline{\gamma} \cdot (p\circ \alpha) \cdot \gamma = p\circ (\overline{\widetilde{\gamma}} \cdot \alpha \cdot \widetilde{\gamma}),\] so indeed we get \([\gamma]^{-1} \cdot p_*([\alpha]) \cdot [\gamma] \in p_*(\pi_1(\widetilde{X},\widetilde{x}'))\) ◻

Proof of (a).. Assume there exists a deck transformation \(f:\widetilde{X},\widetilde{x} \to \widetilde{X},\widetilde{x}'\). Let \(\widetilde{\gamma}\) be any path from \(\widetilde{x}\) to \(\widetilde{x}'\), and denote \(\gamma = p\circ \widetilde{\gamma}\). We show that \([\gamma] \in N(H)\), i.e., for any \(p_*([\alpha]) \in H\), \[[\gamma]^{-1} \cdot p_*([\alpha]) \cdot [\gamma] \in H.\] By lemma 29, we can write \[[\gamma]^{-1} \cdot p_*([\alpha]) \cdot [\gamma] = p_*([\beta])\] for some \(\beta\in \pi_1(\widetilde{X},\widetilde{x}')\) for some \(\widetilde{x}'\) which is another lift of \(x\). Using the fact that \(f\) is a deck transformation, we can insert \(f\) to get \[p_*([\beta]) = p_*(f_*([\beta])),\] so we get \[[\gamma]^{-1} \cdot p_*([\alpha]) \cdot [\gamma] = p_*([\beta]) \in H.\]

Conversely, assume that there exists a path \(\widetilde{\gamma}\) from \(\widetilde{x}\) to \(\widetilde{x}'\) such that for \(\gamma=p\circ \widetilde{\gamma}\), we have \([p\circ \widetilde{\gamma}]\in N(H)\). Then, to get the desired deck transformation, we try to lift \(p:\widetilde{X},\widetilde{x}\to X,x\) to the covering space \(p:\widetilde{X},\widetilde{x}'\to X,x\). There indeed exists a lift \(f:\widetilde{X},\widetilde{x} \to \widetilde{X},\widetilde{x}'\) by the lifting criterion theorem 17 because \[H=p_*(\pi_1(\widetilde{X},\widetilde{x})) \subseteq p_*(\pi_1(\widetilde{X},\widetilde{x}')),\] since by lemma 29, for \(\widetilde{\gamma}\) a path \(\widetilde{x}\) to \(\widetilde{x}'\) and \(\gamma=p\circ \overline{\gamma}\), we have \[p_*(\pi_1(\widetilde{X},\widetilde{x}')) = [\gamma]^{-1} \cdot p_*(\pi_1(\widetilde{X},\widetilde{x})) \cdot [\gamma] = [\gamma]^{-1} \cdot H \cdot [\gamma],\] and by assumption \([\gamma]^{-1}\cdot H \cdot [\gamma] = H\) because \([\gamma]\in N(H)\).

Additionally, this deck transformation is unique, since for another deck transformation \(g\) mapping \(\widetilde{x}\mapsto \widetilde{x}'\), we have that \(g\) is also a lift of \(p:\widetilde{X},\widetilde{x}\to X,x\) to the covering space \(p:\widetilde{X},\widetilde{x}'\to X,x\), since \(p\circ g = p\). Therefore, by uniqueness in proposition 18, we must have \(g=f\). ◻

Proof of (b).. Assume \(p\) is normal. Then, for any \([\gamma]\in \pi_1(X,x)\), denote \(\widetilde{\gamma}\) the lift starting at \(\widetilde{x}\), and let \(\widetilde{x}'=\widetilde{\gamma}(1)\) its other endpoint. Since \(p\) is normal, there is a deck transformation \[f:\widetilde{X},\widetilde{x} \longrightarrow \widetilde{X},\widetilde{x}',\] so by part (a), we get \([\gamma] \in N(H)\). Therefore, \(N(H)=\pi_1(X,x)\) so \(H\trianglelefteq\pi_1(X,x)\).

Conversely, assume that \(H\trianglelefteq\pi_1(X,x)\). A priori, we should show for any \(y\in X\) and two pairs of lifts \(\widetilde{y},\widetilde{y}'\), there exists a deck transformation sending \(\widetilde{y} \mapsto \widetilde{y}'\). However, we can assume \(y=x\), since otherwise we could use lemma 29 to switch to \(y\) as our basepoint, while still getting \(H\) a normal subgroup. Additionally, we can assume \(\widetilde{y}=\widetilde{x}\) because we can compose multiple deck transformations to map between arbitrary lifts of \(x\).

Therefore, we just need to show the existence of a deck transformation mapping \(\widetilde{x}\mapsto \widetilde{x}'\). Let \(\widetilde{\gamma}\) be a path from \(\widetilde{x}\) to \(\widetilde{x}'\). Since \(H\) is a normal subgroup, we have \([p\circ \widetilde{\gamma}] \in N(H)\). Thus, by part (a) there exists the desired deck transformation. ◻

Proof of (c).. We construct a (surjective) group homomorphism \(\varphi: N(H)\to G(\widetilde{X})\), and show the kernel is \(H\). For \([\gamma]\in N(H)\), lift the loop to a path \(\widetilde{\gamma}\) starting at \(\widetilde{x}\) and define \(\widetilde{x}'=\widetilde{\gamma}(1)\). By part (a), there exists a unique deck transformation \(f\) mapping \(\widetilde{x}\mapsto \widetilde{x}'\), so define \(\varphi([\gamma]) = f\).

The function \(\varphi\) is indeed a homomorphism, since \(\varphi([\beta])\circ \varphi([\alpha])\) is still a deck transformation, and it maps \(\widetilde{x}\) to \(\widetilde{\beta\cdot \alpha}(1)\), so we must have \[\varphi([\beta])\circ \varphi([\alpha]) = \varphi([\beta][\alpha]).\]

The surjectivity of \(\varphi\) is immediate from part (a). Finally, \([\gamma]\in \ker{\varphi}\) if and only if \(\varphi([\gamma])\) maps \(\widetilde{x}\mapsto \widetilde{x}\), which occurs exactly when \([\gamma]\) lifts to a loop at \(\widetilde{x}\), i.e., \([\gamma] \in H\).

Thus, by the first isomorphism theorem \[N(H)/H \cong G(\widetilde{X}).\] ◻

Proof of (a).. We first show that \(p\) is a covering map. Let \(G\cdot y \in Y/G\). Choose \(U\) to be a neighborhood of \(y\) as given by the definition of \(G\) being a covering space action. Then, \(G\cdot U \subseteq Y/G\) has preimage \(\{g\cdot U \operatorname{\big|}g\in G\}\) which is a disjoint union of sets homeomorphic to \(U\) (since \(g\) acts as a homeomorphism, and by assumption the \(g\cdot U\) are disjoint). Thus, \(U\) is an evenly covered neighborhood, so \(p\) is a covering map.

Next, we show \(p: Y\to Y/G\) is a normal covering space. For \(y',y\in G\cdot y\), by definition there is a \(g\in G\) such that \(g\cdot y = y'\), and this \(g\) acts as a deck transformation. ◻

Proof of (b).. Identifying \(G\) as a subset of \(\operatorname{Homeo}(Y)\), we can quickly get \(G\subseteq G(Y)\). Conversely, let \(f\in G(Y)\) and \(y\in Y\). Since \(f\) is a deck transformation \(f(y)\in G\cdot y\), so there exists \(g\in G\) such that \(g\cdot y = f(y)\). By proposition 30 (a), \(f=g\) because \(Y\) is path connected and both map \(y\mapsto g\cdot y\). Thus, \(f\in G\), so \(G\cong G(Y)\) by the map that includes \(G\hookrightarrow \operatorname{Homeo}(Y)\). ◻

Proof of (c).. This follows from (a), (b), and proposition 30. ◻

Proof. Suppose, to the contrary, that there is no such \(\delta\). Therefore, for \(\delta=\frac{1}{n}\), there exists \(x_n\in X\) with \(B_{\frac{1}{n}}(x_n)\) not contained in any \(U_i\). By compactness, take a convergent subsequence \(\{x_{n_k}\}\) converging to \(x\in X\). Since \(x\in X\), it is in some open set in the cover, say \(U_{i_0}\). Therefore, there exists \(\delta>0\) such that \(B_{\delta}(x) \subseteq U_{i_0}\).

However, we can choose \(k\in \mathbb{N}\) large so that \[d(x_{n_k}, x) < \frac{\delta}{2}\] and at the same time we can choose \(k\geq \frac{2}{\delta}\) so that \(n_k \geq \frac{2}{\delta}\). By our construction of the sequence, this means that \[B_{\frac{1}{n_k}}(x_{n_k}) \not\subseteq U_{i_0},\] but on the other hand we have \[B_{\frac{1}{n_k}}(x_{n_k}) \subseteq B_{\frac{\delta}{2}}(x_{n_k}) \subseteq B_{\delta}(x)\] because for any \(y\in B_{\frac{\delta}{2}}(x_{n_k})\), \(d(x,y)\leq d(x,x_{n_k}) + d(x_{n_k},y) < \delta\), so \(y\in B_{\delta}(x)\).

However, this gives us \(B_{\frac{1}{n_k}}(x_{n_k}) \subseteq B_{\delta}(x)\subseteq U_{i_0}\), which is a contradiction.

Thus, there does exist a \(\delta>0\) such that every \(\delta\)-ball in \(X\) is contained in some \(U_i\). ◻

Proof. Let \(E\subseteq Y\) be closed. It suffices to show that \(f^{-1}(E)\) is closed. We have \[\begin{aligned} f^{-1}(E) &= f^{-1}(E) \cap (A\cup B)\\ &= (f^{-1}(E)\cap A) \cup (f^{-1}(E)\cap B)\\ &= (f|_A)^{-1}(E) \cup (f|_B)^{-1}(E). \end{aligned}\] Since \(A,B\) are closed, the fact that \[(f|_A)^{-1}|(E), \quad (f|_B)^{-1}|(E)\] are closed in the subspace topologies implies that they are closed in all of \(X\) as well because they are intersections closed subsets of \(X\) with \(A,B\) (which already closed).

(This is the point of the proof—we are using the closed definition of continuity rather than the open definition because of this). ◻

Proof. Let \(V\subseteq Y\) be open. Then, \[\begin{aligned} f^{-1}(V) &= \bigcup_{i\in I} f|_{U_i}^{-1}(V). \end{aligned}\] Each \(f|_{U_i}\) is continuous, so by the subspace topology \[f|_{U_i}^{-1}(V) = W_i \cap U_i\] for some open set \(W_i\). Then, \[f^{-1}(V) = \bigcup_{i\in I} W_i\cap U_i,\] so \(f^{-1}(V)\) is the union of open sets and hence open. Thus, \(f\) is continuous. ◻