$S^1$ is homeomorphic to $[0, 1]/ \{0, 1\}$.

Question

How can I able to show that $S^1$ is homeomorphic to $[0, 1]/ \{0, 1\}.$

I am learning quotient topology from K.D.Joshi's Introduction to GENERAL TOPOLOGY book. Where he mentioned that,

"Let $f: X\to Y$ be a surjective function. Then $f$ determines an equivalence relation $R$ on $X$ defined by $x R y$ iff $f(x) = f(y)$. The equivalence classes of $R$ are precisely the inverse images of singleton subsets of $Y$. Now let $D$ be the collection of all equivalence classes under $R$. $D$ is called the quotient set of $X$ by $R$ and is also denoted by $X/R$.
There is a canonical function $p : X \to X/R$, called the projection which assigns to each $x \in X$, its equivalence class under $R$. The function $\theta : Y \to X/R$ defined by $\theta(y) = p(x)$ for any $x \in f^{-1}(y)$ is obviously a well-defined bijection.

Suppose now that $X, Y$ are topological spaces and that $f$ is a quotient map. On $X/R$, we put the strong topology generated by the projection function $p$. The function $\theta$ then becomes continuous as its composite with $f$ viz., $\theta\circ f$ is continuous. Similarly $\theta^{-1}$ is continuous. Thus $\theta$ is now not merely a bijection but a homeomorphism. Thus, up to a topological equivalence, we may identify the quotient space $Y$ with the quotient space $X/R$ and the quotient map $f$ with the projection $p$.
$S^1$ is homeomorphic to the quotient space of $[0, 1]$ obtained from the
decomposition $D$ whose members are ${0, 1}$ and all singleton sets $\{x\}$ for $0 < x < 1$.

By his notations How can I find $f$ and $\theta$. Can someone help me please? Please explain elaborately since I am very new to the subject and I have no teacher to learn from. Thank you.

That quote is somewhat problematic: it says that the function $\theta$ is a homeomorphism, but it never introduces a topology on $X/R$. Moreover, even when that is done properly, it is not true that the map is always a homeomorphism! For example, the map $t\in[0,1)\mapsto\exp(2\pi i t)\in S^1$ is surjective and the corresponding $\theta$ is not a homeo. (This can be fixed by adding hypotheses: if your quotation is incomplete, it would be best to complete it...) — Mariano Suárez-Álvarez, Jun 29 '13 at 03:46
In any case, do you know of any surjective continuous map $[0,1]\to S^1$? — Mariano Suárez-Álvarez, Jun 29 '13 at 03:49

score 3 · Answer 1 · answered Jun 29 '13 at 09:59

3

The function $f:[0,1] \to S^1$, defined by $f(x) = (\cos(2\pi x), \sin(2\pi x))$, is surjective and continuous and has the property that $f(x) = f(x')$ iff ($x = x'$ or $\{x,x'\} = \{0,1\}$). So the decomposition $R$ from $f$ of $[0,1]$, as in the first paragraph, is exactly your $D$ from below. So here $X = [0,1], Y = S^1$. The $\theta$ from $S^1$ to $X/R$ is just (as defined in the first paragraph) the function that maps a point $(\cos(2\pi x),\sin(2\pi x))$ on the circle to the equivalence class of $x$, where we ensure that $x \in [0,1]$. Any point on the circle can be written that way.

answered Jun 29 '13 at 09:59

Henno Brandsma

242,131

1

I am demonstrating this through the universal property of the quotient topology. I have already proven that $\bar{f}([t])=f(t)$ where $\bar{f}:[0,1]/{0,1}\rightarrow S^1$ is continuous, surjective and injective but I have not been able to demonstrate that it is an open function, how can I do this? Thank you very much. – user425181 Oct 24 '17 at 23:35
@user425181 It's a closed map by compactness and that's enough. – Henno Brandsma Sep 25 '21 at 07:56
Hi Henno, how can we prove that $f(x)=f(x') \Rightarrow x=x' \hbox{ or } {x,x'}={0,1}$? – Math Apr 13 '23 at 02:07

$S^1$ is homeomorphic to $[0, 1]/ \{0, 1\}$.

1 Answers1