[proofplan] The exponential map $p: \mathbb{R} \to S^1$, $t \mapsto e^{2\pi i t}$, is a covering map whose total space $\mathbb{R}$ is simply connected, so $p$ is the universal cover of $S^1$ and the fibre $p^{-1}(1)$ is $\mathbb{Z} \subset \mathbb{R}$. By the [Universal Cover Bijection](/theorems/1893), the map $\ell([\gamma]) := \tilde{\gamma}(1)$ is a bijection $\pi_1(S^1, 1) \to \mathbb{Z}$. The only remaining work is to upgrade bijection to group homomorphism: we exhibit a canonical representative $u_n := p \circ \tilde{u}_n$ for the fibre class labelled $n$ (where $\tilde{u}_n(t) := nt$), use simple connectivity of $\mathbb{R}$ to see every class $[\gamma]$ with $\tilde{\gamma}(1) = n$ equals $[u_n]$, and compute directly that concatenation $u_m \cdot u_n$ lifts to a path from $0$ to $m + n$, giving $\ell([u_m][u_n]) = m + n$. [/proofplan] [step:Identify the exponential map as the universal cover of $S^1$] Define \begin{align*} p: \mathbb{R} &\to S^1 \\ t &\mapsto e^{2\pi i t}. \end{align*} We view $S^1 \subseteq \mathbb{C}$ with basepoint $1 = e^0$, so $p^{-1}(1) = \mathbb{Z} \subset \mathbb{R}$. The map $p$ is a covering map: each $z \in S^1$ has an open neighbourhood $V \subset S^1$ (e.g. a proper open arc) such that $p^{-1}(V)$ is a disjoint union of open intervals in $\mathbb{R}$, each mapped homeomorphically onto $V$ by $p$. The total space $\mathbb{R}$ is path-connected and simply connected (it is convex, hence contractible; see [Characterisation of Simply Connected Spaces](/theorems/1883)). Therefore $p$ is a **universal cover** of $S^1$, with distinguished fibre point $\tilde{x}_0 := 0 \in p^{-1}(1)$. [guided] We verify carefully that the exponential map is a universal covering. First we write the map as a concrete function with its domain and codomain: \begin{align*} p: \mathbb{R} &\to S^1 \\ t &\mapsto (\cos 2\pi t, \sin 2\pi t) = e^{2\pi i t}. \end{align*} With the basepoint convention $x_0 = 1 \in S^1$, the fibre over $1$ consists of exactly those $t \in \mathbb{R}$ with $e^{2\pi i t} = 1$, i.e. $p^{-1}(1) = \mathbb{Z}$. **Why is $p$ a covering map?** For any point $z_0 = e^{2\pi i s_0} \in S^1$, the open arc $V := \{e^{2\pi i s} : s \in (s_0 - 1/4, s_0 + 1/4)\}$ is an open neighbourhood of $z_0$. Its preimage \begin{align*} p^{-1}(V) = \bigsqcup_{n \in \mathbb{Z}} (s_0 - 1/4 + n, \, s_0 + 1/4 + n) \end{align*} is a disjoint union of open intervals, each of length $1/2$, and $p$ restricts to a homeomorphism from each onto $V$ (it is continuous, bijective on each slice because the period of $p$ is $1$, and has continuous inverse given by a branch of $(2\pi i)^{-1} \log$). So $V$ is evenly covered. **Why is $\mathbb{R}$ simply connected?** The space $\mathbb{R}$ is path-connected (the straight-line homotopy $s \mapsto (1 - s)a + sb$ is a path from $a$ to $b$). For the fundamental group: any loop $\gamma: I \to \mathbb{R}$ at $0$ is null-homotopic via the straight-line homotopy $H(s, t) := (1 - t)\gamma(s)$. Hence $\pi_1(\mathbb{R}, 0) = \{e\}$ — see [Characterisation of Simply Connected Spaces](/theorems/1883) for the equivalent formulations. Since $p$ is a covering map with simply connected total space, $p$ is a **universal cover**. Fix the basepoint $\tilde{x}_0 := 0 \in p^{-1}(1)$. [/guided] [/step] [step:Apply the Universal Cover Bijection to obtain a bijection $\ell: \pi_1(S^1, 1) \to \mathbb{Z}$] Apply [Universal Cover Bijection](/theorems/1893) to the universal cover $p: \mathbb{R} \to S^1$ with basepoints $x_0 = 1$, $\tilde{x}_0 = 0$. The hypotheses of that theorem — $p$ a universal cover, $S^1$ path-connected — are satisfied (path-connectedness of $S^1$ holds since $S^1$ is the continuous image of $I$ under $p|_{[0,1]}$). The conclusion gives the bijection \begin{align*} \ell: \pi_1(S^1, 1) &\to p^{-1}(1) = \mathbb{Z} \\ [\gamma] &\mapsto 0 \cdot [\gamma] = \tilde{\gamma}(1), \end{align*} where $\tilde{\gamma}$ is the unique lift of $\gamma$ starting at $0$, provided by the [Path Lifting Lemma](/theorems/1886) and made unique by [Uniqueness of Lifts](/theorems/1885). [guided] Having established that $p: \mathbb{R} \to S^1$ is a universal cover in Step 1, we plug this into [Universal Cover Bijection](/theorems/1893). The theorem demands: (i) $X$ is path-connected — here $X = S^1$, which is path-connected because it is the continuous image of the path-connected interval $[0, 1]$ under $t \mapsto e^{2\pi i t}$; (ii) $p$ is a universal cover — verified in Step 1; (iii) a choice of basepoint $\tilde{x}_0$ in the fibre over $x_0$ — we take $x_0 = 1 \in S^1$ and $\tilde{x}_0 = 0 \in \mathbb{Z} = p^{-1}(1)$. The conclusion of the theorem is that the orbit map \begin{align*} \ell: \pi_1(S^1, 1) &\to p^{-1}(1) = \mathbb{Z} \\ [\gamma] &\mapsto 0 \cdot [\gamma] \end{align*} is a bijection. Unwrapping the action: for a loop $\gamma: I \to S^1$ based at $1$, the [Path Lifting Lemma](/theorems/1886) applied to the covering $p: \mathbb{R} \to S^1$ and initial point $0$ produces a unique lift $\tilde{\gamma}: I \to \mathbb{R}$ with $\tilde{\gamma}(0) = 0$ and $p \circ \tilde{\gamma} = \gamma$; uniqueness follows from [Uniqueness of Lifts](/theorems/1885). Then $0 \cdot [\gamma] = \tilde{\gamma}(1)$, which lies in $p^{-1}(1) = \mathbb{Z}$. Therefore $\ell([\gamma]) = \tilde{\gamma}(1) \in \mathbb{Z}$, exactly matching the theorem statement. *What remains to prove?* Only the homomorphism property $\ell([\gamma] \cdot [\delta]) = \ell([\gamma]) + \ell([\delta])$. [/guided] [/step] [step:Normalise each class by a canonical winding loop $u_n$] For $n \in \mathbb{Z}$, define \begin{align*} \tilde{u}_n: I &\to \mathbb{R} \\ t &\mapsto nt \end{align*} and $u_n := p \circ \tilde{u}_n: I \to S^1$, so $u_n(t) = e^{2\pi i n t}$. Since $\tilde{u}_n(0) = 0$ and $u_n(0) = u_n(1) = 1$, the path $u_n$ is a loop in $S^1$ at $1$ and $\tilde{u}_n$ is its lift to $\mathbb{R}$ starting at $0$, ending at $n$. We show: if $[\gamma] \in \pi_1(S^1, 1)$ satisfies $\tilde{\gamma}(1) = n$, then $[\gamma] = [u_n]$ in $\pi_1(S^1, 1)$. Indeed, $\tilde{\gamma}$ and $\tilde{u}_n$ are both paths in $\mathbb{R}$ from $0$ to $n$. Define the straight-line homotopy \begin{align*} \tilde{H}: I \times I &\to \mathbb{R} \\ (t, s) &\mapsto (1 - s)\tilde{\gamma}(t) + s \tilde{u}_n(t). \end{align*} This is continuous, satisfies $\tilde{H}(t, 0) = \tilde{\gamma}(t)$, $\tilde{H}(t, 1) = \tilde{u}_n(t)$, and fixes the endpoints $\tilde{H}(0, s) = 0$, $\tilde{H}(1, s) = n$ for all $s$. Composing with $p$ yields a path homotopy $H := p \circ \tilde{H}: I \times I \to S^1$ from $\gamma$ to $u_n$ rel endpoints. Therefore $[\gamma] = [u_n]$. [guided] The goal of this step is to produce, for each integer $n$, a concrete loop $u_n$ in $S^1$ whose lift ends at $n$, and to show this loop represents the *unique* homotopy class in $\ell^{-1}(n)$. **Construction.** The simplest path in $\mathbb{R}$ from $0$ to $n$ is the linear one. Define \begin{align*} \tilde{u}_n: I \to \mathbb{R}, \qquad t \mapsto nt. \end{align*} Push down via $p$: \begin{align*} u_n := p \circ \tilde{u}_n: I \to S^1, \qquad t \mapsto e^{2\pi i n t}. \end{align*} Since $\tilde{u}_n(0) = 0$, $\tilde{u}_n(1) = n$, and $p(\mathbb{Z}) = \{1\}$, we have $u_n(0) = u_n(1) = 1$, so $u_n$ is a loop at $1$. By [Uniqueness of Lifts](/theorems/1885), $\tilde{u}_n$ is *the* unique lift of $u_n$ starting at $0$, and $\ell([u_n]) = \tilde{u}_n(1) = n$. **Canonicality.** Suppose $[\gamma] \in \pi_1(S^1, 1)$ is any class with $\ell([\gamma]) = n$, i.e. whose lift $\tilde{\gamma}$ from $0$ ends at $n$. We must show $[\gamma] = [u_n]$. Both $\tilde{\gamma}$ and $\tilde{u}_n$ are paths in $\mathbb{R}$ from $0$ to $n$. The space $\mathbb{R}$ is convex, so any two paths with the same endpoints are homotopic rel endpoints via the straight-line homotopy \begin{align*} \tilde{H}(t, s) := (1 - s)\tilde{\gamma}(t) + s \tilde{u}_n(t). \end{align*} Check the boundary conditions: $\tilde{H}(t, 0) = \tilde{\gamma}(t)$, $\tilde{H}(t, 1) = \tilde{u}_n(t)$, $\tilde{H}(0, s) = (1-s)\cdot 0 + s \cdot 0 = 0$, and $\tilde{H}(1, s) = (1-s) n + s n = n$. Continuity is immediate from the formula. Now project the homotopy down via the covering map: \begin{align*} H: I \times I \to S^1, \qquad H(t, s) := p(\tilde{H}(t, s)) = e^{2\pi i \tilde{H}(t, s)}. \end{align*} Continuity of $H$ follows from continuity of $p$ and $\tilde{H}$. The boundary conditions become $H(t, 0) = p(\tilde{\gamma}(t)) = \gamma(t)$, $H(t, 1) = p(\tilde{u}_n(t)) = u_n(t)$, and $H(0, s) = p(0) = 1 = H(1, s)$ for all $s$. So $H$ is a path homotopy from $\gamma$ to $u_n$ rel endpoints, giving $[\gamma] = [u_n]$ in $\pi_1(S^1, 1)$. *Why does this matter?* It means we can compute $\ell$ on the canonical representatives $[u_n]$ and extend to all of $\pi_1(S^1, 1)$ by homotopy invariance. This is exactly what the next step does. [/guided] [/step] [step:Compute the lift of the concatenation $u_m \cdot u_n$] Fix $m, n \in \mathbb{Z}$. Recall the concatenation of paths is defined by \begin{align*} u_m \cdot u_n: I \to S^1, \qquad t \mapsto \begin{cases} u_m(2t), & 0 \leq t \leq 1/2, \\ u_n(2t - 1), & 1/2 \leq t \leq 1. \end{cases} \end{align*} Define a candidate lift \begin{align*} \tilde{\sigma}_{m,n}: I &\to \mathbb{R} \\ t &\mapsto \begin{cases} \tilde{u}_m(2t) = 2mt, & 0 \leq t \leq 1/2, \\ \tilde{u}_n(2t - 1) + m = n(2t-1) + m, & 1/2 \leq t \leq 1. \end{cases} \end{align*} We verify three properties of $\tilde{\sigma}_{m,n}$. *Continuity:* The two branches agree at $t = 1/2$: the first gives $2m \cdot 1/2 = m$, the second gives $n(0) + m = m$. Continuity on each closed half-interval follows from continuity of $\tilde{u}_m, \tilde{u}_n$. By the [Gluing Lemma](/theorems/1871) applied to the closed cover $I = [0, 1/2] \cup [1/2, 1]$, $\tilde{\sigma}_{m,n}$ is continuous on $I$. *Projection:* On $[0, 1/2]$, $p(\tilde{\sigma}_{m,n}(t)) = p(2mt) = e^{2\pi i \cdot 2mt} = u_m(2t) = (u_m \cdot u_n)(t)$. On $[1/2, 1]$, using $p(x + m) = p(x)$ for $m \in \mathbb{Z}$, \begin{align*} p(\tilde{\sigma}_{m,n}(t)) = p(n(2t-1) + m) = p(n(2t-1)) = u_n(2t - 1) = (u_m \cdot u_n)(t). \end{align*} *Initial point:* $\tilde{\sigma}_{m,n}(0) = 2m \cdot 0 = 0$. By [Uniqueness of Lifts](/theorems/1885), $\tilde{\sigma}_{m,n}$ is *the* lift of $u_m \cdot u_n$ starting at $0$. Its terminal value is \begin{align*} \tilde{\sigma}_{m,n}(1) = n(2 \cdot 1 - 1) + m = n + m. \end{align*} [guided] To check that $\ell$ is a homomorphism, we need to evaluate $\ell([u_m] \cdot [u_n]) = \ell([u_m \cdot u_n])$. By Step 2, this is the terminal value of the unique lift of $u_m \cdot u_n$ starting at $0$. We construct that lift explicitly. **Strategy.** The lift of $u_m$ starting at $0$ is $\tilde{u}_m$, ending at $m$. After $u_m$ is traversed, the concatenated path $u_n$ begins at $1 \in S^1$; its lift should start *where the previous lift ended*, namely at $m \in \mathbb{R}$. But a lift of $u_n$ starting at $m$ is just $\tilde{u}_n$ translated by $m$: the map $t \mapsto \tilde{u}_n(t) + m = nt + m$ still projects to $u_n$ (because $p(x + m) = p(x)$ for integer $m$) and begins at $m$. So the candidate lift, reparametrised to match the domain of $u_m \cdot u_n$, is \begin{align*} \tilde{\sigma}_{m,n}(t) = \begin{cases} 2mt, & 0 \leq t \leq 1/2, \\ n(2t - 1) + m, & 1/2 \leq t \leq 1. \end{cases} \end{align*} **Verification.** 1. *Agreement at $t = 1/2$:* From the left, $2m \cdot 1/2 = m$. From the right, $n(1 - 1) + m = m$. The two formulas agree, so [Gluing Lemma](/theorems/1871) (applied to the closed sets $[0, 1/2]$ and $[1/2, 1]$ covering $I$) gives continuity of $\tilde{\sigma}_{m,n}$ on $I$. 2. *Projection $p \circ \tilde{\sigma}_{m,n} = u_m \cdot u_n$:* On $[0, 1/2]$, $p(2mt) = e^{4\pi i mt} = u_m(2t)$. On $[1/2, 1]$, using that $p$ has period $1$, \begin{align*} p(n(2t - 1) + m) = e^{2\pi i (n(2t-1) + m)} = e^{2\pi i n (2t - 1)} \cdot e^{2 \pi i m} = e^{2\pi i n (2t-1)} = u_n(2t - 1). \end{align*} These match the two pieces of the concatenation $u_m \cdot u_n$. 3. *Initial point:* $\tilde{\sigma}_{m,n}(0) = 0$. By [Uniqueness of Lifts](/theorems/1885), applied to the connected interval $I$, $\tilde{\sigma}_{m,n}$ is *the* unique lift of $u_m \cdot u_n$ starting at $0$. Its terminal value is \begin{align*} \tilde{\sigma}_{m,n}(1) = n(2 - 1) + m = m + n, \end{align*} so $\ell([u_m \cdot u_n]) = m + n$. [/guided] [/step] [step:Conclude $\ell$ is an isomorphism] Combining the previous steps, for any $m, n \in \mathbb{Z}$, \begin{align*} \ell([u_m] \cdot [u_n]) = \ell([u_m \cdot u_n]) = \tilde{\sigma}_{m,n}(1) = m + n = \ell([u_m]) + \ell([u_n]). \end{align*} By Step 3, every class $[\gamma] \in \pi_1(S^1, 1)$ is of the form $[u_n]$ for the unique $n = \ell([\gamma]) \in \mathbb{Z}$, so the identity above determines $\ell$ on all of $\pi_1(S^1, 1)$ as a group homomorphism. Since $\ell$ is already a bijection by Step 2, it is a group isomorphism \begin{align*} \ell: \pi_1(S^1, 1) \xrightarrow{\sim} \mathbb{Z}. \end{align*} This completes the proof. [/step]