$\mathbb{R} / \mathbb{Z}$ is homeomorphic to $S^{1}$ .

Question

I have been attending some seminars at my university where I am a 1st year PhD student.

In one of the seminars my colleague left these questions as trivial but I am having a really hard time proving them. Actually, my topology is not so good.

Questions:

(a) Consider $\mathbb{R} /\mathbb{Z}$ equipped with quotient topology. Show that $\mathbb{R} / \mathbb{Z}$ is homeomorphic to $S^1 \subset \mathbb{R}^2$.

(b) Consider $[0,1]$ equipped with the equivalence relation $x \sim y \Leftrightarrow x=y$ or $x= 1$ and $y=0$ or $x=0$ and $y=1$. Show that $[0,1]/ \sim$ equipped with quotient topology is homeomorphic to $S^1$.

Attempt: (a) I have to find a continuous map from $\mathbb{R} / \mathbb{Z}$ to $S^1$ such that it's inverse is also continuous.$ f(x)= e^{2\pi ix}$ came to my mind and I thought it as an map from $\mathbb{R} \to S^1$. It's inverse is = $\log x/ (2\pi i)$ which is continuous . But how can I find a map from $\mathbb{R} / \mathbb{Z}$ instead of $\mathbb{R}$. I am not able to think about it.

(b) So, I have been following Wayne Patty's Foundations of Topology and in the book, definition of Quotient topology and all related theorems is given in terms of functions and because of that reason I am not able to prove this part(b) or make any progress. Can you please let me know whic results should I use to solve this part?

Thanks!

Answer 1

These questions has been asked many times here, but I was not able to find references in math.stackexchange with really complete answers. Perhaps somebody else will find something.

So let us do it again.

(a) You considered the map $f : \mathbb R \to S^1, f(x) = e^{2\pi i x}$. This is the right approach. In fact it is well-known that

1. $f$ is a surjective group homomorphism from the additive group $(\mathbb R, +)$ to the multiplicative group $(S^1, \cdot)$ of complex numbers with absolute value $1$.

2. The kernel of $f$ is the subgroup $\mathbb Z \subset \mathbb R$. In other words, $f(x) = f(y)$ iff $x - y \in \mathbb Z$.

$\mathbb R/\mathbb Z$ denotes the quotient of $\mathbb R$ modulo the subgroup $\mathbb Z$. Its elements are the cosets of $\mathbb Z$ in $\mathbb R$, i.e. the sets $[x]_{\mathbb Z} = x + \mathbb Z = \{x +k \mid k \in \mathbb Z\}$ with $x \in \mathbb R$. We have $[x]_{\mathbb Z} = [y]_{\mathbb Z}$ if and only if $x - y \in \mathbb Z$. Alternatively one can regard the elements of $\mathbb R/\mathbb Z$ as the equivalence classes $[x]_\sim$ of the points $x \in \mathbb R$ with respect to the equivalence relation defined by $x \sim y$ iff $x - y \in \mathbb Z$. Clearly $[x]_{\mathbb Z} = [x]_\sim$.

Let $p : \mathbb R \to \mathbb R/\mathbb Z, p(x) = [x]_{\mathbb Z}$, denote the canonical quotient map. This is both a topological quotient map to the quotient space $\mathbb R/\mathbb Z$ which is endowed with the quotient topology and an algebraic quotient map (i.e. a group homomorphism) to the quotient group $\mathbb R/\mathbb Z$.

$f$ induces a unique function $\bar f : \mathbb R/\mathbb Z \to S^1$ such that $\bar f \circ p = f$. It is well-known that $\bar f$ is continuous and a group isomorphism $\mathbb R/\mathbb Z = \mathbb R/\ker f \to \operatorname{im} f = S^1$. It remains to show that $\bar f$ is a homeomorphism. We know that it is continuous bijection. But clearly $\mathbb R/\mathbb Z = p([0,1])$, thus $\mathbb R/\mathbb Z$ is the continuous image of a compact set and therefore itself compact. This implies that $\bar f$ is a homeomorphism because $S^1$ is Hausdorff.

(b) Let $r : X \to Y$ be a quotient map between spaces $X,Y$. Define an equivalence relation $\sim_r$ on $X$ by $x \sim_r x'$ iff $r(x) = r(x')$. It is well-known that the induced map $\bar r : X/\sim_r \to Y$ (characterized by $\bar r \circ q = r$, where $q : X \to X/\sim_r$ is the standard quotient map) is a homeomorphism.

Now let $r = p \mid_{[0,1]} : [0,1] \to S^1$. This is a continuous surjection. Since $[0,1]$ is compact and $S^1$ is Hausdorff, it is a closed map and therefore a quotient map. The equivalence relation $\sim_r$ agrees with your $\sim$. Thus $\bar r : [0,1]/\sim \phantom . \to S^1$ is a homeomorphism.

Remark:

You could also use the method in (b) to prove (a). But it is somewhat more difficult to prove that $f$ is a quotient map. See here where it is shown that $f$ is an open map and therefore a quotient map.

Answer 2

(a) Let $S^1$ be the unit circle on the complex plane. Define $\phi: \Bbb{R}/\Bbb{Z}\rightarrow S^1$ by the rule $\phi([x])=e^{2\pi ix}$ for every $[x]\in\Bbb{R}/\Bbb{Z}.$ This function is well-defined, since if $x'$ is another representetive of the class $[x]$ then $x'=x+n$ for some $n\in\Bbb{Z}$ and $$\phi([x'])=e^{2\pi ix'}=e^{2\pi i(x+n)}=e^{2\pi ix}=\phi([x]).$$ This function is continuous. To prove this it is enough to prove that the pre-image of the basis elements in the topology of $S^1$ are open. These basis elements are open arcs. The inverse image of such an arc is open because, the arc contains 1 or not, it is in the form $\pi((a,b))$ for some $a,b\in\Bbb{R}$ where $\pi: \Bbb{R}\rightarrow \Bbb{R}/\Bbb{Z}$ is the quotient map which is an open map. This function is clearly surjective. It is injective, because if $\phi([x_1])=\phi([x_2])$, then $x_1=x_2+n$ for some integer $n$ and hence $[x_1]=[x_2]$. Hence, $\phi$ has an inverse $\phi^{-1}: S^1\rightarrow \Bbb{R}/\Bbb{Z}$. The inverse is also continuous, because basis elements of the topology of $\Bbb{R}/\Bbb{Z}$ are $\pi((a,b))$ and their pre-images are open arcs.

(b) Define $\psi: I/\sim\rightarrow S^1$ by the rule $\psi([x])=e^{2\pi ix}$ for every $[x]\in I/\sim .$ This function is well-defined because $\psi(0)=\psi(1)=1$. $\psi$ is continuus because, the pre-image of an open arc is in the form $(a,b)/\sim$ if it does not contain $1\in S^1$; in the form $[0,a)\cup(b,1]/\sim$, if it contains $1$ for some $0<a,b<1$ in $\Bbb{R}$ and these are open sets in that quotient topology. $\psi$ is clearly injective and surjective. It has an inverse. $\psi^{-1}$ is continuous.To see this consider again the basis element of the topology of $I/\sim$. (Warning: Subsets like $[0,a)$ or $(b,1]$ are open sets of $I$, but their image are not open in $I/\sim$. That is the quotient map $\pi: I\rightarrow I/\sim$ is not open.)