Understanding a composition mapping to a quotient space

Question

If we have the following atlas for the circle $S^1$ $$\{(\{e^{it}\in\mathbb{C}|-\pi<t<\pi\},\phi_1(e^{it})=t),(\{e^{it}\in\mathbb{C}|0<t<2\pi\},\phi_2(e^{it})=t)\}$$ and let $\bar{\phi}_i$ be $\phi_i$ followed by the projection $\mathbb{R}\to \mathbb{R}/2\pi\mathbb{Z}$. On $U_1\cap U_2$, $\phi_1$ and $\phi_2$ differ by an integer multiple of $2\pi$, so $\bar{\phi}_1=\bar{\phi}_2$, and we piece them together to form $\bar{\phi}:S^1\to\mathbb{R}/2\pi\mathbb{Z}$, which is then well defined.
This is the setup for exercise 7.7 in Tu's "Introduction to Manifolds". I have been trying to understand how to represent the map $\bar{\phi}$, but it is not entirely clear to me how to think of $\mathbb{R}/2\pi\mathbb{Z}$. Judging by some other online resources, it seems that we can identify it with the interval $[0,2\pi)$, in which case, would it be correct to say that $\bar{\phi}(e^{it})=t\mod{2\pi}$?

Answer 1

The purpose of the exercise is to show that $F : \mathbb R/2\pi \mathbb Z \to S^1$ is a is diffeomorphism. This is why Tu introduces $\bar \phi : S^1 \to \mathbb R/2\pi \mathbb Z$ - this map is inverse to $F$.

You may think of $\mathbb R/2\pi \mathbb Z$ as the interval $[0,2\pi]$ with endpoints identified. In fact, let $i : [0,2\pi] \to \mathbb R$ denote inclusion. Then $q = p \circ i : [0,2\pi] \to \mathbb R/2\pi \mathbb Z$ is a continuous surjection because each $x \in \mathbb R$ is equivalent under the action of $2\pi \mathbb Z$ to a number $x' \in [0,2\pi]$. Moreover, $q(x) = q(x')$ iff $x=x'$ or $x,x' \in \{0,2\pi\}$. This means that $q$ induces a homeomorphism $Q: [0,2\pi]/\{0,2\pi\} \to \mathbb R/2\pi \mathbb Z$. But clearly $[0,2\pi]/\{0,2\pi\}$ is homeomorphic to $S^1$. Hence you must not think that we can identify $\mathbb R/2\pi \mathbb Z$ with the interval $[0,2π)$. But it is correct that $$\bar \phi (e^{it}) = t \mod 2 \pi .$$

The essence is that each $z \in S^1$ can be written as $z = e^{it}$, but $t$ is determined only up to an integral multiple of $2\pi$. That is the reason for working $\mod 2 \pi$.

Of course we can represent each $z \in S^1$ as $z = e^{it}$ with a unique $t \in [0,2π)$. But as said above, this does not mean that we can identify $\mathbb R/2\pi \mathbb Z$ with $[0,2π)$. We only have a continuous bijection $[0,2π) \to \mathbb R/2\pi \mathbb Z$, but it is no homeomorphism.