I have just begun learning algebraic topology, and was trying to find an elementary proof of the fact that the fundamental group of $S^1$ is isomorphic to $\mathbb{Z}$. I think I now have an answer, but since I am unable to find this approach online (perhaps a consequence of not knowing the terminology/keywords pertaining to the lemma involved), I was hoping to get it cross-checked. In addition, I was also hoping to know spaces to which I could generalise the following lemma (if at all it is correct as it currently stands). Thank you!
Lemma : Let $A \subseteq \mathbb{R}^N$ be a closed rectangle and $f : A \to S^1$ be a continuous function. Then, upto addition by $2 n \pi$ for $n \in \mathbb{Z}$, there exists a continuous parametrization $\theta : A \to \mathbb{R}$ such that $f = e^{i\theta}$.
Proof : The uniqueness part of the proof follows from the fact that the difference between any two parametrizations is an integer multiple of $2\pi$ at each point, and that the image of a connected set under a continuous function is connected.
For the existence part, I first explicitly construct $\theta$ in the cases where the image of $\theta$ is nice:
- $\theta = \sin^{-1} \circ \Im f$, when $f\left(A\right) \subseteq S^1 \cap \{z \in \mathbb{C} \, : \, \Re(z) > 0 \} = U_1$
- $\theta = \cos^{-1} \circ \Re f$, when $f\left(A\right) \subseteq S^1 \cap \{z \in \mathbb{C} \, : \, \Im(z) > 0 \} = U_2$
- $\theta = \pi - \sin^{-1} \circ \Im f$, when $f\left(A\right) \subseteq S^1 \cap \{z \in \mathbb{C} \, : \, \Re(z) < 0 \} = U_3$
- $\theta = - \cos^{-1} \circ \Re f$, when $f\left(A\right) \subseteq S^1 \cap \{z \in \mathbb{C} \, : \, \Im(z) < 0 \} = U_4$
Now for the general case, we can assume WLOG that $A = [0, 1]^N$. For $n \in \{1, \ldots, N\}$, let $P(n)$ := for all $x_n, x_{n + 1}, \ldots, x_N \in [0, 1]$, there exist intervals $\{I_n, \ldots, I_N\}$ open in $[0, 1]$ such that $x_i \in I_i$ and there exists a parametrization $\theta$ on $[0, 1]^{n - 1} \times \Pi_{i = n}^N I_i(= \Pi_{i = 1}^N I_i$ when $n = 1$). Let $P(N + 1)$ claim the existence of a parametrization on $A$.
Let $y \in A$. Since $S^1 = \bigcup_{i = 1}^4 U_i$, it follows that $y \in f^{-1}(U_i)$ for some $i \in \{1, 2, 3, 4\}$. Taking the restriction of the corresponding parametrization $\theta$ to a basic open set (in the product topology) containing $y$, $P(1)$ is proven.
Assume $P(n)$ for some $n \in \{1, \ldots N\}$. If $n < N$, fix $x_{n + 1}, \ldots, x_N$. In the following steps, I will subsume the case $n = N$ by assuming the convention of ignoring products of the form $\Pi_{i = m}^{m'} I_i$ when $m > m'$.
Let \begin{align*} \mathcal{O} = \{[0, 1]^{n - 1} \times \Pi_{i = n}^N I_i \, : \, I_n, \ldots, I_N \text{ are intervals open in $[0, 1]$ such that there exists a parametrization $\theta$ on } [0, 1]^{n - 1} \times \Pi_{i = n}^N I_i\} \end{align*}
Note that if $[0, 1]^{n - 1} \times \Pi_{i = n}^N I_i \in \mathcal{O}$, then by taking the restriction of the corresponding parametrization, it follows that $[0, 1]^{n - 1} \times \Pi_{i = n}^N I_i' \in \mathcal{O}$ for every $I'_i \subseteq I_i$.
It follows from $P(n)$ that $\mathcal{O}$ is non-empty and covers $[0, 1]^n \times \Pi_{i = n + 1}^N \{x_i\}$. Since the aforementioned set is compact, there exists a finite subcover $\{[0, 1]^{n - 1} \times \Pi_{i = n}^N {I_j}_i\}_{j = 1}^{k'}$. It follows that $\mathcal{A} = \{[0, 1]^{n - 1} \times I_j \times V\}_{j = 1}^{k'}$ is also a subcover of $\mathcal{O}$, where $I_j = {I_j}_n$ and $V = \Pi_{i = n + 1}^N (\bigcap_{l = 1}^{k'} {I_l}_i)$ (not applicable when $n = N$, ignore $V$ for said case henceforth). Consider a minimal subcover $\mathcal{B}$ of $\mathcal{A}$. WLOG assume $|\mathcal{B}| = k > 1$ (else we are already done). Then, there exists an enumeration $\mathcal{B} = \{[0, 1]^{n - 1} \times I_j \times V\}_{j = 1}^k$ such that $I_1 = [a_1, b_1)$, $I_k = (a_k, b_k]$, $I_j = (a_j, b_j)$ for $I \in \{2, \ldots, k - 1\}$ satisfying
\begin{align*} 0 = a_1 < a_2 < b_1 < a_3 < b_2 < \ldots < a_k < b_{k - 1} < b_k = 1 \end{align*}
Set $W_i = [0, 1]^{n - 1} \times I_j \times V$. By the nature of this enumeration, $\bigcup_{i = 1}^l W_i$ and $(\bigcup_{i = 1}^l W_i) \cap W_{l + 1}$ are closed cubes for $l \in \{1, \ldots k - 1\}$. Assume that there exists a parametrization $\theta'$ on $\bigcup_{i = 1}^l W_i$ (this is true for $l = 1$ since $W_1 \in \mathcal{O})$. Let $\theta_{W_{l + 1}}$ be a parametrization on $W_{l + 1}$. Then, by uniqueness, $\theta'$ and $\theta_{W_{l + 1}}$ differ on $(\bigcup_{i = 1}^l W_i) \cap W_{l + 1}$ by $2n\pi$ for some $n \in \mathbb{Z}$. By the gluing lemma, $\theta$ defined to be $\theta'$ on $\bigcup_{i = 1}^l W_i$ and $\theta_{W_{l + 1}} + 2n\pi$ on $W_{l + 1}$ is a parametrization on $\bigcup_{i = 1}^{l + 1} W_i$. Hence, there exists a parametrization $\theta$ on $[0, 1]^n \times \Pi_{i = n + 1}^N \{x_i\} = \bigcup_{i = 1}^k W_i$, implying $P(n + 1)$.
Therefore, by induction, $P(N + 1)$ is true. $$\tag*{$\blacksquare$}$$
Consider the map $\tilde{\phi} : \bigcup \pi(S^1, 1) \to \mathbb{Z}$ as follows :
\begin{align*} \tilde{\phi}\left(e^{i\theta}\right) = \dfrac{\theta(1) - \theta(0)}{2\pi} \end{align*}
From the above lemma, it follows that this map is well defined. Let $\tilde{\phi}\left(e^{i\theta}\right) = \tilde{\phi}\left(e^{i\theta'}\right)$. WLOG assume $\theta(0) = \theta'(0)$ and $\theta(1) = \theta'(1)$ Then, the map defined by $(s, t) \mapsto e^{i\left( s\theta'(t) + (1 - s)\theta(t)\right)}$ is a path homotopy from $e^{i\theta}$ to $e^{i\theta'}$. Conversely, let $\gamma_1, \gamma_2 \in \bigcup \pi(S^1, 1)$ be path homotopic. Let $H : [0, 1]^2 \to S^1$ be a path homotopy from $\gamma_1$ to $\gamma_2$. Then, there exists a function $\theta : [0, 1]^2 \to \mathbb{R}$ such that $H = e^{i\theta}$. It follows that $t \mapsto \theta(i, t)$ is a parametrization for $\gamma_i$. For $j \in \{0, 1\}$ , consider the function $f_j : [0, 1] \to \mathbb{R}$ defined by $f_j(s) = \dfrac{\theta(s, j) - \theta(0, j)}{2\pi}$. Since $H$ is a path homotopy, it follows from elementary trigonometry that $f_j$ is integer valued, and since $f_j(0) = 0$, it follows from the continuity of $f_j$ and the connectedness of $[0, 1]$ that $f_j = 0$. In particular, $\theta(1, j) = \theta(0, j)$ for $I \in \{0, 1\}$. Hence, $\tilde{\phi}(\gamma_1) = \tilde{\phi}(\gamma_2)$.
Therefore, the function $\phi : \pi_1(S^1, 1) \to \mathbb{Z}$ defined by $\phi([\gamma]) = \tilde{\phi}(\gamma)$ is a well defined bijection. It remains to show that $\phi$ is a group homomorphism.
Let $[e^{i\theta_1}], [e^{i\theta_2}] \in \pi_1(S^1, 1)$. WLOG assume that $\theta_1(1) = \theta_2(0)$. Then, the function $\theta$ defined below is a parametrization of the concatenation $e^{i\theta_1} \cdot e^{i\theta_2}$.
\begin{align*} \theta = \begin{cases} \theta_1(2s) &, s \in \left[0, \dfrac{1}{2}\right]\\ \theta_2(2s - 1) &, s \in \left[\dfrac{1}{2}, 1\right] \end{cases} \end{align*}
Then,
\begin{align*} \phi([e^{i\theta_1}][e^{i\theta_2}]) & = \phi([e^{i\theta_1} \cdot e^{i\theta_2}])\\ & = \tilde{\phi}(e^{i\theta_1} \cdot e^{i\theta_2})\\ & = \tilde{\phi}(e^{i\theta})\\ & = \dfrac{\theta(1) - \theta(0)}{2\pi}\\ & = \dfrac{\theta_2(1) - \theta_1(0)}{2\pi}\\ & = \dfrac{\theta_1(1) - \theta_1(0)}{2\pi} + \dfrac{\theta_2(1) - \theta_2(0)}{2\pi}\\ & = \tilde{\phi}(e^{i\theta_1}) + \tilde{\phi}(e^{i\theta_2})\\ & = \phi([e^{i\theta_1}]) + \phi([e^{i\theta_2}])\\ \end{align*}
Therefore, $\phi$ is a group isomorphism.
