Fundamental Group of the Circle

The computation $\pi_1(S^1) \cong \mathbb{Z}$ is the first nontrivial calculation in algebraic topology and serves as the foundation for many further results. The proof uses the covering space $\mathbb{R} \to S^1$ and the path lifting property.

Statement

Theorem9.9Fundamental Group of $S^1$

The fundamental group of the circle is infinite cyclic: $\pi_1(S^1, 1) \cong \mathbb{Z}.$

The isomorphism sends $n \in \mathbb{Z}$ to the homotopy class of the loop $\omega_n(t) = e^{2\pi i n t}$ , which winds $n$ times around the circle (counterclockwise for $n > 0$ , clockwise for $n < 0$ ).

Proof Outline

Proof

Let $p: \mathbb{R} \to S^1$ be the covering map $p(s) = e^{2\pi i s}$ , with $p(0) = 1$ .

Step 1: Define the degree map.

For a loop $\gamma: [0, 1] \to S^1$ based at $1$ , let $\tilde{\gamma}$ be the unique lift to $\mathbb{R}$ with $\tilde{\gamma}(0) = 0$ . Since $p(\tilde{\gamma}(1)) = \gamma(1) = 1$ , we have $\tilde{\gamma}(1) \in p^{-1}(1) = \mathbb{Z}$ .

Define $\deg: \pi_1(S^1, 1) \to \mathbb{Z}$ by $\deg([\gamma]) = \tilde{\gamma}(1)$ .

Step 2: Well-definedness.

If $\gamma_0 \simeq \gamma_1$ (rel $\{0, 1\}$ ) via homotopy $H$ , then $H$ lifts to $\tilde{H}: [0,1]^2 \to \mathbb{R}$ . The path $t \mapsto \tilde{H}(1, t)$ is a continuous function from $[0, 1]$ to the discrete set $\mathbb{Z}$ , hence constant. So $\tilde{\gamma}_0(1) = \tilde{H}(1, 0) = \tilde{H}(1, 1) = \tilde{\gamma}_1(1)$ .

Step 3: $\deg$ is a homomorphism.

If $[\gamma_1], [\gamma_2] \in \pi_1(S^1, 1)$ with lifts $\tilde{\gamma}_1$ and $\tilde{\gamma}_2$ , the lift of $\gamma_1 * \gamma_2$ is $\tilde{\gamma}_1$ followed by the translate $\tilde{\gamma}_2(\cdot) + \tilde{\gamma}_1(1)$ . The endpoint is $\tilde{\gamma}_1(1) + \tilde{\gamma}_2(1)$ . So $\deg([\gamma_1] \cdot [\gamma_2]) = \deg[\gamma_1] + \deg[\gamma_2]$ .

Step 4: $\deg$ is surjective.

The loop $\omega_n(t) = e^{2\pi i n t}$ lifts to $\tilde{\omega}_n(t) = nt$ , which has $\tilde{\omega}_n(1) = n$ . So $\deg([\omega_n]) = n$ .

Step 5: $\deg$ is injective.

Suppose $\deg([\gamma]) = 0$ , i.e., $\tilde{\gamma}(1) = 0 = \tilde{\gamma}(0)$ . Then $\tilde{\gamma}$ is a loop in $\mathbb{R}$ based at $0$ . Since $\mathbb{R}$ is simply connected ( $\pi_1(\mathbb{R}) = 0$ ), $\tilde{\gamma}$ is null-homotopic: there exists $\tilde{H}: \tilde{\gamma} \simeq \tilde{e}$ (rel $\{0, 1\}$ ). Then $H = p \circ \tilde{H}$ is a homotopy $\gamma = p \circ \tilde{\gamma} \simeq p \circ \tilde{e} = e$ (rel $\{0, 1\}$ ). So $[\gamma] = [e]$ .

Therefore $\deg: \pi_1(S^1, 1) \to \mathbb{Z}$ is an isomorphism.

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Key Lemmas Used

RemarkIngredients of the Proof

The proof relies on three key facts about the covering $p: \mathbb{R} \to S^1$ :

Path lifting: Every path in $S^1$ lifts uniquely to $\mathbb{R}$ (given a starting point).
Homotopy lifting: Path homotopies lift to $\mathbb{R}$ , ensuring well-definedness.
Simple connectivity of $\mathbb{R}$ : This ensures injectivity of the degree map.

The general principle: the fundamental group $\pi_1(X)$ acts on the fiber $p^{-1}(x_0)$ of the universal cover by deck transformations, and this action identifies $\pi_1(X)$ with the group of deck transformations. For $S^1$ , the deck transformations of $\mathbb{R} \to S^1$ are the translations $t \mapsto t + n$ for $n \in \mathbb{Z}$ .

Consequences

Theorem9.10$S^1$ is Not Simply Connected

The circle $S^1$ is not simply connected. The loop $\omega_1(t) = e^{2\pi i t}$ generates $\pi_1(S^1)$ and is not null-homotopic.

ExampleThe Winding Number

For a continuous map $f: S^1 \to S^1$ , the degree of $f$ (or winding number) is $\deg(f) = n$ where $f_*([\omega_1]) = [\omega_n]$ , i.e., $f_*: \mathbb{Z} \to \mathbb{Z}$ is multiplication by $n$ .

$\deg(\operatorname{id}) = 1$ .
$\deg(z \mapsto z^n) = n$ .
$\deg(\text{constant}) = 0$ .
$\deg(z \mapsto \bar{z}) = -1$ (complex conjugation reverses orientation).

Two maps $f, g: S^1 \to S^1$ are homotopic if and only if $\deg(f) = \deg(g)$ .

ExampleNo Retraction from $D^2$ to $S^1$

There is no continuous retraction $r: D^2 \to S^1$ (i.e., no continuous map $r$ with $r \circ \iota = \operatorname{id}_{S^1}$ where $\iota: S^1 \hookrightarrow D^2$ is the inclusion).

If $r$ existed, then $r_* \circ \iota_*: \pi_1(S^1) \to \pi_1(S^1)$ would be the identity. But $\iota_*: \pi_1(S^1) \to \pi_1(D^2)$ maps $\mathbb{Z} \to 0$ (since $D^2$ is contractible), so $r_* \circ \iota_*$ is the zero map. Contradiction.