Answer: The Covering Space That Knows Its Sheets

Key Idea / Intuition

The deep insight is that the fiber $p^{-1}(x)$ is in natural bijection with the coset space $\pi_1(X,x) / p_*\pi_1(\widetilde{X}, \widetilde{x})$. The bijection works by lifting loops: two loops at $x$ land at the same point in the fiber if and only if one composed with the reverse of the other lifts to a loop (rather than an open path) in $\widetilde{X}$, which is exactly the condition for their classes to lie in the same left coset of $p_*\pi_1(\widetilde{X}, \widetilde{x})$. So the fiber is "indexed by cosets," and the number of sheets literally counts how many cosets there are.

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Formal Proof / Solution

Setup. Fix a basepoint $x \in X$ and $\widetilde{x} \in p^{-1}(x)$. Write $H = p_* \pi_1(\widetilde{X}, \widetilde{x}) \leq \pi_1(X, x)$.

Step 1: Define a map from left cosets to the fiber.

For any loop $\gamma$ based at $x$, the unique lift $\widetilde{\gamma}$ starting at $\widetilde{x}$ ends at some point $\widetilde{\gamma}(1) \in p^{-1}(x)$. Define

$\Phi: \pi_1(X,x)/H \;\longrightarrow\; p^{-1}(x), \qquad [\gamma] \cdot H \;\mapsto\; \widetilde{\gamma}(1).$

Step 2: Well-definedness. If $[\gamma'] \in [\gamma] \cdot H$, then $[\gamma' \cdot \gamma^{-1}] \in H = p_* \pi_1(\widetilde{X},\widetilde{x})$, so $\gamma' \cdot \gamma^{-1}$ lifts to a loop in $\widetilde{X}$ based at $\widetilde{x}$. By uniqueness of path lifting, the lift of $\gamma'$ also ends at $\widetilde{\gamma}(1)$.

Step 3: Injectivity. If $\widetilde{\gamma}(1) = \widetilde{\gamma'}(1)$, then $\widetilde{\gamma}^{-1} \cdot \widetilde{\gamma'}$ is a loop in $\widetilde{X}$ based at $\widetilde{x}$, so $[\gamma^{-1} \cdot \gamma'] \in H$, meaning $[\gamma]$ and $[\gamma']$ are in the same left coset.

Step 4: Surjectivity. Since $\widetilde{X}$ is path-connected, for any $\widetilde{y} \in p^{-1}(x)$ there is a path $\widetilde{\alpha}$ from $\widetilde{x}$ to $\widetilde{y}$ in $\widetilde{X}$. Then $\gamma = p \circ \widetilde{\alpha}$ is a loop at $x$, and $\Phi([\gamma] H) = \widetilde{y}$.

Conclusion. $\Phi$ is a bijection, so

$\#\, p^{-1}(x) \;=\; [\pi_1(X,x) : H]. \qquad \square$

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The $n$-sheeted cover $p(z) = z^n$ on $S^1$.

We know $\pi_1(S^1, 1) \cong \mathbb{Z}$, generated by the loop $\omega(t) = e^{2\pi i t}$.

The map $p_*: \pi_1(S^1, 1) \to \pi_1(S^1, 1)$ sends $[\omega]$ to $[p \circ \omega]$. Since $p(\omega(t)) = e^{2\pi i n t}$, this is the loop that winds around $n$ times, so $p_*([\omega]) = n[\omega]$.

Therefore:

$p_* \pi_1(S^1, 1) \;=\; n\mathbb{Z} \;\leq\; \mathbb{Z},$

and the index is

$[\mathbb{Z} : n\mathbb{Z}] = n,$

which exactly matches the $n$ sheets $\{1, e^{2\pi i/n}, e^{4\pi i/n}, \ldots, e^{2\pi i(n-1)/n}\}$ in the fiber over $1$.

Elegant summary table:

| $n$-sheeted cover | Subgroup of $\mathbb{Z}$ | Index | |---|---|---| | $p(z) = z^n$ | $n\mathbb{Z}$ | $n$ |

This shows the Galois correspondence for covering spaces in a concrete form: the subgroups of $\mathbb{Z}$ are exactly $n\mathbb{Z}$ for $n \geq 0$, and each corresponds to exactly one (connected) covering of $S^1$ up to isomorphism.