Answer: The Covering Space That Wraps Around Twice

Key Idea / Intuition

The map $z \mapsto z^2$ wraps the circle around itself twice — every point has exactly two preimages, and small arcs upstairs map homeomorphically to arcs downstairs. The fundamental group of $S^1$ is $\mathbb{Z}$, generated by the loop that goes around once. Upstairs, going around once only covers half the circle downstairs, so the induced map on $\pi_1$ must be multiplication by $2$. The general pattern reveals that every nontrivial subgroup of $\mathbb{Z}$ (all of which are of the form $n\mathbb{Z}$) appears as the image of such a covering map.

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Formal Proof / Solution

Part (a): $p(z) = z^2$ is a 2-sheeted covering map

We need to show every point of $S^1$ has an evenly covered neighborhood.

Given any $w \in S^1$, let $U$ be an open arc in $S^1$ not containing $-w$ (the antipodal point). Then: $p^{-1}(U) = V_1 \sqcup V_2$ where $V_1$ and $V_2$ are the two open arcs of $S^1$ mapping to $U$ — specifically the two square roots of points in $U$. These arcs are disjoint (since $-w \notin U$ means the two arcs don't overlap), and $p\big|_{V_i}: V_i \to U$ is a homeomorphism for each $i$.

Thus $p$ is a covering map with 2 sheets (each fiber $p^{-1}(w)$ has exactly 2 points: $\{w^{1/2}, -w^{1/2}\}$).

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Part (b): The induced map $p_*: \mathbb{Z} \to \mathbb{Z}$

Recall $\pi_1(S^1, 1) \cong \mathbb{Z}$, generated by the loop $\gamma: [0,1] \to S^1, \quad \gamma(t) = e^{2\pi i t},$ which winds around once (representing $1 \in \mathbb{Z}$).

The induced map sends $[\gamma] \mapsto [p \circ \gamma]$, and: $(p \circ \gamma)(t) = \gamma(t)^2 = e^{4\pi i t}.$

This loop winds around $S^1$ twice, so $[p \circ \gamma] = 2 \in \mathbb{Z}$.

Therefore: $\boxed{p_*: \mathbb{Z} \to \mathbb{Z}, \quad p_*(n) = 2n.}$

That is, $p_*$ is multiplication by $2$.

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Part (c): All subgroups of $\mathbb{Z}$ via covering maps

For the $n$-sheeted cover $p_n(z) = z^n$, the same argument gives: $(p_n \circ \gamma)(t) = e^{2\pi i n t},$ which winds around $n$ times, so $(p_n)_*: \mathbb{Z} \to \mathbb{Z}$ is multiplication by $n$.

The image of $(p_n)_*$ is the subgroup $n\mathbb{Z} \leq \mathbb{Z}$.

Now recall the classification theorem: the subgroups of $\mathbb{Z}$ are exactly $\{0\} = \{0\}$ and $n\mathbb{Z}$ for $n \geq 1$. These are in bijection with the covering spaces of $S^1$:

| Cover | Map | $(p_n)_*$ image | Subgroup | |---|---|---|---| | $S^1 \xrightarrow{z^n} S^1$ | $n$-sheeted | multiplication by $n$ | $n\mathbb{Z}$ | | $\mathbb{R} \to S^1$ (universal) | $\infty$-sheeted | trivial | $\{0\}$ |

Conclusion: Every subgroup of $\mathbb{Z}$ arises as the image of the induced map on $\pi_1$ from some covering map $S^1 \to S^1$. This is a perfect illustration of the correspondence between subgroups of $\pi_1(B)$ and covering spaces of $B$: the image $p_*(\pi_1(E))$ inside $\pi_1(B)$ completely characterizes the covering up to equivalence.

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Bonus elegance: The trivial subgroup $\{0\}$ corresponds to the universal cover $\mathbb{R} \to S^1$, $t \mapsto e^{2\pi i t}$, where $\pi_1(\mathbb{R}) = 0$. The whole group $\mathbb{Z}$ (the identity cover) corresponds to $z \mapsto z$, a 1-sheeted cover.