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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?

Martin Sleziak
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Sharabh
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3 Answers3

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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).

Michael Albanese
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  • Ohh.. so we are actualy using two homeomorphisms : first from R/Z to [0,1) and then from [0,1) to S1. Cool! Thanx. – Sharabh Sep 17 '13 at 01:21
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    @Sharabh: Yes, but you could compose them to get the map $x + \mathbb{Z} \mapsto e^{2\pi i x}$. Now you can see that it doesn't really matter which representative you choose for $x + \mathbb{Z}$. – Michael Albanese Sep 17 '13 at 01:26
  • and one more thing. when you are writing 10.9+Z as 0.9+Z, do you mean they belong to the same class ? – Sharabh Sep 17 '13 at 01:28
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    I mean that $10.9 + \mathbb{Z} = 0.9 + \mathbb{Z}$. Equivalently, $10.9$ and $0.9$ belong to the same equivalence class under the equivalence relation $x \sim y$ if $x - y \in \mathbb{Z}$. – Michael Albanese Sep 17 '13 at 01:29
  • oh.. ok ok. Got it. Thanks. – Sharabh Sep 17 '13 at 01:31
  • No, [0,1) is not homeomorphic to the two other spaces. There is a continuous bijection from [0,1) to the circle, but it’s Inverse map is not continuous. [0,1) is not compact after all. – user39082 Jun 17 '21 at 15:30
  • @user39082: You're thinking about the subspace topology, but as I mentioned in my answer, that is not the topology that $[0, 1)$ is equipped with in order to be homeomorphic to $S^1$. – Michael Albanese Jun 17 '21 at 15:48
  • @MichaelAlbanese Is the topology on $[0,1)$ generated by sets of form $(a,b), 0 \leq a < b \leq 1$ and $[0,a) \cup (b,1), 0 < a < b < 1$ – somitra Apr 02 '23 at 07:54
  • @somitra: That is correct. – Michael Albanese Apr 02 '23 at 14:13
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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 $f=\widetilde{f}\circ p$.

Xiang Yu
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  • What is the best way to show that it is indeed a quotient map? Is there a trick we can use to skip the technical details? – Dean Gurvitz Feb 02 '19 at 11:37
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    $p$ is surjective and open (check that each open interval in $\mathbb{R}$ is mapped to an open subset of $S^1$). It is a well-known fact that open surjections are quotient maps. – Paul Frost Feb 08 '19 at 11:08
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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.

André Nicolas
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