2

Hi please could someone check my understanding here please? I've been asked to find a quotient map from $\mathbb R$ to $S^1$.

Is the quotient map the surjective function from $\mathbb R$ to $\mathbb R/\sim$ where an open set in $\mathbb R/\sim$ is open iff it's pre-image is open in $\mathbb R$, or is it the surjective function from $\mathbb R$ to $S^1$ where an open set in $S^1$ is open iff it's pre-image is open in $\mathbb R$?

I've been using my lecture notes and the book by W A Sutherland but I'm struggling to follow the phrasing for questions such as this.

TIA

Jake Andrews
  • 47

1 Answers1

1

Let's consider the map $\exp_I:I\to S^1$ such that $t\mapsto(\cos(2\pi t),\sin (2\pi t))=e^{2\pi it}$.
First we want to show that $\exp_I$ is an identification. So, in particular we have to prove that the topology on $S^1$ is the same topology induced by $\exp_I$.
Firts we can observe that $\exp_I$ is continuous because $\exp:\mathbb R\to \mathbb R^2$ is continuous.
We call $\mathcal T_d$ the topology on $S^1$ induced by the euclidean distance and we'll show that $\mathcal T_d=\mathcal T_{\exp_I}$.
The open saturated sets in $[0,1]$ are like $[0,a)\cup(b,1]$ or $(c,d)$, with $0<a<b<1$ and $0<c<d<1$. Notice that $\exp_I^{-1}(\exp_I([0,a)))=[0,a)\cup \{1\}$, which is not open.
The open sets in $(S^1,\exp_I)$ are given by the images of the sets in $\{[0,a)\cup(b,1],(c,d)\}\overset{a,b,c,d\in \mathbb R}{\underset{0<c<d<1}{_{0<a<b<1}}}$, which are arches on $S^1$. Since open sets of $\mathcal T_d$ are given by the intersection of $\epsilon$-balls in $\mathbb R^2$ with the circle $S^1$, we have shown that $\mathcal T_{\exp_I}<\mathcal T_d$.
Conversely is quite easy because open sets for $\mathcal T_d$ are the sets $B_{\epsilon}(p)\cap S^1$ that are open for $\mathcal T_{\exp_I}$ and this shows that $\mathcal T_d<\mathcal T_{\exp_I}\implies \mathcal T_d=\mathcal T_{\exp_I}\implies\exp_I$ is an identification and we can use the universal property of quotient.

We put on $I$ the topology induced by the euclidean topology on $\mathbb R$, so $A\subseteq I$ is open if $\exists U \subseteq \mathbb R$ open such that $A=U\cap I$.
On $S^1$ we consider the topology induced by the map $\exp_I$, so a set $A$ is open for $\mathcal T_{\exp_I}\iff \exp^{-1}(\exp_I(A))=A$.

We now define an equivalence relation $\sim$ such that $$t\sim t'\iff\begin{cases}t=t'\\t,t'\in \{0,1\}\end{cases}\iff\exp_I(t)=\exp_I(t').$$ Let $\pi:I\to I/\sim$ be the quotient map, whose topology on $I/\{0,1\}$ is the topology induced by the map $\pi$.

Since $\exp_I$ is an identification and respect the property $[t]_{\sim}=[t']_{\sim}\implies \exp_I(t)=\exp_I(t')$, we say that $\bar{\exp_I}:I/\{0,1\}\to S^1$ is well-defined, hence injective. Furthermore $\bar{\exp_I}$ is surjective because $\exp_I$ is surjective and it's also continuous for the universal property of quotient $\implies\bar{\exp_I}$ is an homeomorphism whose inverse is $\bar \pi:S^1\to I/\{0,1\}$.
We can conclude that $(I/\{0,1\},\mathcal T_{\pi})\cong(S^1,\mathcal T_d)$.

Vajra
  • 2,802
  • I hope this can help you. – Vajra Dec 31 '20 at 16:57
  • Thank you @Vajra. So to confirm we would say that $\mathbb{R} \rightarrow \mathbb{R} / $~ is the quotient map and that $\mathbb{R} \rightarrow S^1$ is just a map? – Jake Andrews Dec 31 '20 at 17:03
  • It's not clear from this why $\bar{\text{exp}}_I$ is a homeomorphism. – Jakobian Dec 31 '20 at 17:04
  • He showed confusion in the construction in general. If he wants I could add all the details of the proof. – Vajra Dec 31 '20 at 17:06
  • Thanks for this. I think is enough detail for my course. I'm merely confused as it stands on which function is called the quotient map for the question 'Define a quotient map from $\mathbb{R}$ to $S^1$'. Is it $\mathbb{R} \rightarrow \mathbb{R}$/~ or $\mathbb{R} \rightarrow S^1$? – Jake Andrews Dec 31 '20 at 17:07
  • I didn't mean the details. Just the explanation is lacking at some crucial points. – Jakobian Dec 31 '20 at 17:07
  • I think I understand now. Thank you both! – Jake Andrews Dec 31 '20 at 17:09
  • The quotient map is supposed to be a function $\mathbb{R}\to S^1$. – Jakobian Dec 31 '20 at 17:10
  • I should add that exp is an identification map. This is really important. – Vajra Dec 31 '20 at 17:10
  • In literature, quotient map doesn't restrict to maps $X\to X/\sim$ @Vajra, I don't think that's what they ask for here. – Jakobian Dec 31 '20 at 17:12
  • So when doing the homeomorphism glueing part we introduce a different function from points in $\mathbb{R}$ to their equivalence classes and call this space Y say. Then the function from Y to $S^1$ can be proven to be a homeomorphism by Vajra's argument above? – Jake Andrews Dec 31 '20 at 17:13
  • Apologies for a million questions! Thank you for being patient and explaining thoroughly. I'm reasonably new to topology and have been struggling with this unit quite a bit as is obvious. – Jake Andrews Dec 31 '20 at 17:14
  • You can take the map $\mathbb{R} \to \mathbb{R}/\sim$ and then from $\mathbb{R}\to S^1$ obtain an induced map $\mathbb{R}/\sim \to S^1$, but you won't know if it's a homeomorphism without adressing further details about the nature of the map $\mathbb{R}\to S^1$. – Jakobian Dec 31 '20 at 17:18
  • What is an induced map? Check for homeomorphism in usual way, continuous bijective function with continuous inverse? – Jake Andrews Dec 31 '20 at 17:20
  • This doesn't even matter for the purpose of the question you were trying to originally ask. – Jakobian Dec 31 '20 at 17:20
  • Apologies. Thank you – Jake Andrews Dec 31 '20 at 17:21
  • I mean the unique map $\mathbb{R}/\sim \to S^1$ such that $\mathbb{R}\to S^1$ decomposes into $\mathbb{R}\to \mathbb{R}/\sim \to S^1$. – Jakobian Dec 31 '20 at 17:21
  • I tried to be as clear as possible. – Vajra Dec 31 '20 at 17:52
  • Uhh now I understand!! We started talking about $S^1$ and the quotient space and I missed the original point of the question. I hope my effort won't be useless... Sorry, my fault :-( – Vajra Dec 31 '20 at 17:59