Answer: Fundamental Group of the Circle Is ℤ

Key Idea / Intuition

The real line $\mathbb{R}$ is the "unrolled" version of $S^1$: the covering map $\varepsilon(t) = e^{2\pi i t}$ wraps $\mathbb{R}$ around $S^1$ like a helix over a circle. Every loop in $S^1$ based at $1$ lifts uniquely to a path in $\mathbb{R}$ starting at $0$, and the endpoint of that lifted path must be an integer (since it also maps to $1 \in S^1$). That integer — the winding number — is a homotopy invariant because homotopies lift to homotopies, and a continuous deformation can't jump the endpoint from one integer to another. This single integer completely classifies loops, giving $\pi_1(S^1) \cong \mathbb{Z}$.

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Formal Proof / Solution

Setup: Lifting paths to $\mathbb{R}$

The covering map $\varepsilon : \mathbb{R} \to S^1$, $\varepsilon(t) = e^{2\pi i t}$, has the path lifting property: for every path $\gamma$ in $S^1$ with $\gamma(0) = 1$ there is a unique lift $\tilde{\gamma} : [0,1] \to \mathbb{R}$ with $\tilde{\gamma}(0) = 0$ such that $\varepsilon \circ \tilde{\gamma} = \gamma$.

Since $\gamma$ is a loop ($\gamma(1) = 1$), we need $\varepsilon(\tilde{\gamma}(1)) = 1$, i.e., $e^{2\pi i \tilde{\gamma}(1)} = 1$, so $\tilde{\gamma}(1) \in \mathbb{Z}$.

Definition. The winding number of $\gamma$ is $n(\gamma) := \tilde{\gamma}(1) \in \mathbb{Z}$.

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Part (a): $\gamma_1 \not\simeq \gamma_2$

The lift of $\gamma_n(s) = e^{2\pi i n s}$ starting at $0$ is simply $\tilde{\gamma}_n(s) = ns$, so

$\tilde{\gamma}_1(1) = 1, \qquad \tilde{\gamma}_2(1) = 2.$

Claim: If $\gamma \simeq \gamma'$ (homotopy of based loops), then $n(\gamma) = n(\gamma')$.

Proof of claim: Let $H : [0,1] \times [0,1] \to S^1$ be a based homotopy ($H(s,0) = \gamma$, $H(s,1) = \gamma'$, $H(0,t) = H(1,t) = 1$). By the homotopy lifting property of covering spaces, $H$ lifts to $\tilde{H} : [0,1]^2 \to \mathbb{R}$ with $\tilde{H}(0,0) = 0$.

• Since $H(0,t) = 1$ for all $t$, the path $t \mapsto \tilde{H}(0,t)$ is a lift of the constant loop at $1$ starting at $0$; by uniqueness it equals $0$ for all $t$.
• Since $H(1,t) = 1$ for all $t$, the path $t \mapsto \tilde{H}(1,t)$ is a lift of the constant loop at $1$ starting at $\tilde{H}(1,0) = n(\gamma) \in \mathbb{Z}$; by uniqueness it equals $n(\gamma)$ for all $t$.
• In particular, $\tilde{H}(1,1) = n(\gamma)$, but $\tilde{H}(1,1)$ is also the endpoint of the lift of $\gamma'$, so $n(\gamma') = n(\gamma)$. $\square$

Since $n(\gamma_1) = 1 \neq 2 = n(\gamma_2)$, the loops $\gamma_1$ and $\gamma_2$ are not homotopic.

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Part (b): $\pi_1(S^1, 1) \cong \mathbb{Z}$

Define the map

$\Phi : \pi_1(S^1, 1) \to \mathbb{Z}, \qquad [\gamma] \mapsto n(\gamma).$

Well-defined & injective: Shown above — the winding number is a homotopy invariant, and two loops with equal winding numbers have lifts with the same endpoints.

For injectivity: if $n(\gamma) = n(\gamma')$, then $\tilde{\gamma}$ and $\tilde{\gamma}'$ are paths in $\mathbb{R}$ from $0$ to the same integer $n$. Since $\mathbb{R}$ is simply connected (contractible), there is a homotopy $\tilde{H}$ between them in $\mathbb{R}$, and $\varepsilon \circ \tilde{H}$ descends to a homotopy between $\gamma$ and $\gamma'$ in $S^1$.

Surjective: The loop $\gamma_n(s) = e^{2\pi i ns}$ has winding number $n$, so every integer is achieved.

Homomorphism: Concatenation of loops corresponds to addition of winding numbers: $n(\gamma * \gamma') = n(\gamma) + n(\gamma'),$ because the lift of $\gamma * \gamma'$ travels from $0$ to $n(\gamma)$, then continues to $n(\gamma) + n(\gamma')$.

Hence $\Phi$ is a group isomorphism, and

$\boxed{\pi_1(S^1, 1) \cong \mathbb{Z}.}$

Geometric meaning: The integer $n(\gamma)$ is the winding number — how many times (and in which direction) the loop wraps around the circle. Counterclockwise counts as $+1$, clockwise as $-1$.