How does the quotient $\mathbb{R}/\mathbb{Z}$ become the circle $S^1$?

I came to know that the circle $S^1$ is actually the quotient space $\mathbb{R}/\mathbb{Z}$. But I don't understand how. To my knowledge elements of the quotient space $X/Y$ are of the form $xY$, i.e. the cosets. Right? But how $\mathbb{R}/\mathbb{Z}$ becomes a circle! Can anyone explain?

Solution 1:

You are correct, elements of $\mathbb{R}/\mathbb{Z}$ are the cosets of $\mathbb{Z}$ in $\mathbb{R}$, but as the group structure is additive, one usually writes a coset as $x + \mathbb{Z}$. So $\mathbb{R}/\mathbb{Z} = \{x + \mathbb{Z} \mid x \in \mathbb{R}\}$.

Of course, there is a lot of repetition here as $x + \mathbb{Z} = y + \mathbb{Z}$ whenever $x - y \in \mathbb{Z}$ (the converse is also true). To overcome this, let's try to pick out an element from each coset which will act as a distinguished representative.

One way of doing this is by choosing the smallest non-negative element of each coset; note, each coset has non-negative elements but they are all discrete (in the topological sense), so there is a minimum such element. So for example, for the coset $10.9 + \mathbb{Z}$, the smallest non-negative element is $0.9$, so we instead write the coset as $0.9 + \mathbb{Z}$. Using these distinguished representatives, we can now write the quotient as $\mathbb{R}/\mathbb{Z} = \{x + \mathbb{Z} \mid x \in [0, 1)\}$.

In particular, $\mathbb{R}/\mathbb{Z}$ is in $1-1$ correspondence with the set $[0, 1)$. The group structure also carries over to $[0, 1)$, you can check it is addition modulo $1$. To be more precise, you can endow $[0, 1)$ with that group structure which turns the map $x + \mathbb{Z} \mapsto x$ into an isomorphism.

Finally, we have an isomorphism between $[0, 1)$ and $S^1$, the set of unit length complex numbers, given by $x \mapsto e^{2\pi i x}$.

Furthermore, we can show all three spaces are the same from a topological point of view. That is, $\mathbb{R}/\mathbb{Z}$, $[0, 1)$, and $S^1$ are all homeomorphic (with the appropriate topologies; in particular, the topology on $[0, 1)$ is not the subspace topology).

Solution 2:

It suffices to show the map $p:\mathbb{R}\to S^1$ defined by $p(x)=e^{2\pi ix}$ is a quotient map. Then you can apply the following theorem (see Munkres' Topology Theorem 22.1) to conclude that $\mathbb{R}/\mathbb{Z}$ and $S^1$ are homeomorphic.

Theorem. Let $f:X\to Y$ be a quotient map. Let $\sim$ be a equivalence relation on $X$ by declaring $x\sim x'$ iff $f(x)=f(x')$. Let $X^*$ be the collection of equivalence class of $X$ (i.e., $X^*=\{[x]:x\in X\}$). We give $X^*$ the quotient topology induced by the map $p:X\to X^*$ that sends each point of $X$ to the equivalence class containing it (i.e., $p(x)=[x]$). Then the map $f$ induces a homeomorphism $\widetilde{f}:X^*\to Y$ characterized by $f=\widetilde{f}\circ p$.

Solution 3:

You are right, in principle they are cosets.

However, every coset has a unique representative in the interval $[0,1)$, so any coset can be identified with its representative in $[0,1)$.

Now take a number $\delta\lt 1$, but with $\delta$ very close to $1$. By adding the (equivalence class) of the very small number $1-\delta$ to (the equivalence class of) $\delta$, we get (the equivalence class of) $0$. Thus $\delta$ is very close to $0$. By the identification of $0$ and $1$, we have in effect turned $[0,1)$ into a circle.

This kind of construction is very common in mathematics. For instance, the set of integers is defined in terms of equivalence classes of ordered pairs of natural numbers. But very quickly the fact that the number $-17$ is forgotten.

Similarly, the formal definition of $\mathbb{Z_n}$ is as a certain collection o cosets. But soon we think of it as the numbers $0$ to $n-1$ with a "different" addition and multiplication.

Added: The following is a more formal approach. Define a mapping of $\mathbb{R}/\mathbb{Z}$ to the numbers of the form $e^{it}$ by the map $\phi$ that takes the coset of $x$ to $e^{2\pi i x}$. The points $e^{2\pi i x}$ can be identified with the unit circle.

The mapping $\phi$ is independent of the choice of representative. For if $x$ and $x'$ belong to the same coset, then $e^{2\pi i x}=e^{2\pi i x'}$, since $e^{2\pi i n}=1$ for any integer $n$.

The map $\phi$ is a group isomorphism. With some work, one can show that it is a homeomorphism.