It is known that if \(p:\widetilde X\to X\) is a covering space and \(x_0\in X\) then the cardinality of \(p^{-1}(x_0)\) is the index of \(p_*\pi_1(\widetilde X,y_0)\) in \(\pi_1(X,x_0)\) where \(p(y_0)=x_0\). Use this fact to deduce that there is no \(n\)-sheeted covering of the circle \(S^1\) for any finite \(n\).
Solution
The requested conclusion is false.
1. For every integer \(n\ge1\), the map\[ p_n:S^1\to S^1, \qquad p_n(z)=z^n, \]is an \(n\)-sheeted covering map.
Fix \[ w=e^{i\theta_0}\in S^1. \] Choose \(0<\varepsilon<\pi\) and let \[ U=\{e^{i\theta}:|\theta-\theta_0|<\varepsilon\}, \] where the interval is understood through the chosen local argument around \(w\). For \(k=0,\ldots,n-1\), define \[ U_k = \left\{ e^{i\phi}: \left|\phi-\frac{\theta_0+2\pi k}{n}\right|<\frac{\varepsilon}{n} \right\}. \] These are pairwise disjoint open arcs, and \[ p_n^{-1}(U)=U_0\sqcup\cdots\sqcup U_{n-1}. \] On each \(U_k\), the map \(p_n\) is a homeomorphism onto \(U\), with inverse \[ e^{i\theta} \longmapsto e^{i(\theta+2\pi k)/n} \] using the same local argument on \(U\). Thus \(U\) is evenly covered and has exactly \(n\) sheets. Since \(w\) was arbitrary, \(p_n\) is an \(n\)-sheeted covering.
2. Under the standard identification\[ \pi_1(S^1,1)\cong\mathbb Z, \]the induced homomorphism is\[ (p_n)_*:\mathbb Z\to\mathbb Z, \qquad m\longmapsto nm. \]
A generator of \(\pi_1(S^1,1)\) is represented by \[ \alpha(t)=e^{2\pi i t}. \] Its image under \(p_n\) is \[ (p_n\circ\alpha)(t) =e^{2\pi i n t}, \] which winds \(n\) times around the circle and therefore represents \(n\in\mathbb Z\). Since \((p_n)_*\) is a group homomorphism, it sends \(m\) to \(nm\).
3. Hence\[ (p_n)_*\pi_1(S^1,1)=n\mathbb Z \]and\[ [\mathbb Z:n\mathbb Z]=n. \]
2. The quotient\[ \mathbb Z/n\mathbb Z \]has exactly \(n\) elements, so the subgroup \(n\mathbb Z\) has index \(n\) in \(\mathbb Z\).
4. The quoted covering-space fact is therefore consistent with these covers and contradicts the requested conclusion.
By 1, every fiber of \(p_n\) has cardinality \(n\). By 3, the induced subgroup has index \(n\). Thus the equality quoted in the problem reads \[ |p_n^{-1}(1)| =n =[\pi_1(S^1):(p_n)_*\pi_1(S^1)]. \] So the theorem does not rule out finite-sheeted coverings of \(S^1\); the maps \(p_n(z)=z^n\) provide one for every finite \(n\ge1\).