Why is the fundamental group of the circle π₁(S¹) ≅ ℤ?

The map $\Phi: \pi_1(S^1, 1) \to \mathbb{Z}$ defined by $\Phi([\gamma]) = \tilde{\gamma}(1)$ where $\tilde{\gamma}$ is the unique lift of $\gamma$ to $\mathbb{R}$ with $\tilde{\gamma}(0) = 0$ is indeed a group isomorphism. Here's the formal proof. **Well-definedness:** If $\gamma_1 \sim \gamma_2$ via a homotopy $H: [0,1]^2 \to S^1$, then by the homotopy lifting property, $H$ lifts uniquely to $\tilde{H}: [0,1]^2 \to \mathbb{R}$ with $\tilde{H}(0,\cdot) = 0$. Since $\tilde{H}(1,\cdot)$ is a continuous map $[0,1] \to p^{-1}(1) = \mathbb{Z}$, it must be constant. Hence $\tilde{\gamma}_1(1) = \tilde{\gamma}_2(1)$. **Homomorphism property:** Let $\gamma_1, \gamma_2$ be loops based at $1$. Their concatenation is $$(\gamma_1 \cdot \gamma_2)(t) = \begin{cases} \gamma_1(2t) & 0 \leq t \leq 1/2 \\ \gamma_2(2t-1) & 1/2 \leq t \leq 1 \end{cases}.$$ The lift of $\gamma_1 \cdot \gamma_2$ is $\tilde{\gamma}_1(2t)$ for $t \in [0,1/2]$ and $\tilde{\gamma}_1(1) + \tilde{\gamma}_2(2t-1)$ for $t \in [1/2,1]$. Evaluating at $t=1$: $$\Phi([\gamma_1 \cdot \gamma_2]) = \tilde{\gamma}_1(1) + \tilde{\gamma}_2(1) = \Phi([\gamma_1]) + \Phi([\gamma_2]).$$ **Injectivity:** If $\Phi([\gamma]) = 0$, then $\tilde{\gamma}(1) = 0$, so $\tilde{\gamma}$ is a loop in $\mathbb{R}$. Since $\mathbb{R}$ is contractible, $\tilde{\gamma} \sim \text{const}$, and projecting via $p$ gives $\gamma \sim \text{const}$. **Surjectivity:** For $n \in \mathbb{Z}$, let $\tilde{\gamma}_n(t) = nt$ and set $\gamma_n = p \circ \tilde{\gamma}_n$. Then $\Phi([\gamma_n]) = n$.