Why Is the Trick for Open Sets in Quotient Topology Valid?

[Korybut]

TL;DR Summary

Open covering of Mobius Strip

Hello!
Reading a textbook I found that authors use the same trick to show that subsets of quotient topology are open. And I don't understand why this trick is valid. Below I provide there example for manifold (Mobius strip) where this trick was used

Quote from "Differential Geometry and Mathematical Physics" by Rudolph and Schmidt
Let $M$ be the topological quotient of the open subset $\mathbb{R} \times (-1,1)\subset \mathbb{R}^2$ by the equivalence relation
$$ (s_1,t_1) \sim (s_2,t_2)\;\; \text{iff} \;\; (s_2,t_2)=(s_1+2\pi k,(-1)^k t_1)\;\; \text{for some} \; k \in \mathbb{Z}.$$

$M$ is called the Mobius strip. Let $p : \mathbb{R}\times (-1,1) \rightarrow M$ denote the natural projection. As the quotient of a second countable space, $M$ is second countable. It is Hausdorff: for $m_1,\; m_2 \in M$, define
$$d(m_1,m_2)=\inf\{ \sqrt{(s_2-s_1)^2+(t_2-t_1)^2} :(s_i,t_i)\in p^{-1}(m_i),\; i=1,2\}$$
and show that $d$ is a metric on $M$, compatible with the quotient topology. To construct an atlas, we show that $p$ is open. For every $k\in \mathbb{Z}$, the mapping
$$\phi_k : \mathbb{R}\times (-1,1)\rightarrow \mathbb{R} \times (-1,1),\;\; \phi_k(s,t):=(s+2\pi k, (-1)^k t),$$
is a homemorphism. If $O\subset \mathbb{R}\times (-1,1)$ is open then $\phi_k(O)$ and hence
$$ p^{-1} \big(p(O)\big)=\cup_{k\in \mathbb{Z}} \phi_k(O)$$
is open. Therefore, $p(O)$ is open.

I have several related questions
1. "show that $d$ is a metric on $M$, compatible with the quotient topology." I have to show that metric all the properties on the quotient? Like triangle inequality, non-degeneracy, etc.
2. "To construct an atlas, we show that $p$ is open." How to understand "$p$ is open"? What is the formal definition?
3. At the end authors showed that $p^{-1}(p(O))$ is open and it is obviously enough for them to state that $p(O)$ is open as a consequence. For generic continuous map this is not true and one can easily come up with contr-example. Why does it work here?

P.S. Please, forgive my ignorance. I am physicist who decided to finally figure out several topological question.

 

[fresh_42]

I have several related questions
1. "show that $d$ is a metric on $M$, compatible with the quotient topology." I have to show that metric all the properties on the quotient? Like triangle inequality, non-degeneracy, etc.

Yes, plus that the distance does not depend on the representative of an equivalence class.

2. "To construct an atlas, we show that $p$ is open." How to understand "$p$ is open"? What is the formal definition?

A function is open if it maps open sets to open sets. Given any $O\subseteq \mathbb{R}\times (-1,1)$ conclude that $p(O)\subseteq M$ is open.

3. At the end authors showed that $p^{-1}(p(O))$ is open and it is obviously enough for them to state that $p(O)$ is open as a consequence. For generic continuous map this is not true and one can easily come up with contr-example. Why does it work here?

This is because of the definition of the quotient topology on $M=\mathbb{R}\times (-1,1)/\sim_p$
$$
\mathcal{T}=\{U\subseteq M\,|\,p^{-1}(U)\subseteq \mathbb{R}\times (-1,1)\text{ is open}\}
$$

 

[Korybut]

This is because of the definition of the quotient topology on $M=\mathbb{R} \times (-1,1)/\sim_p$ $$\mathcal{T}=\{U\subseteq M\vert p^{-1}(U)\subseteq \mathbb{R} \times (-1,1)\; \text{is open}\}$$

I am slightly confused. What kind of $U$s are considered here? If open, then I completely fine with this since canonical projection is continuous by definition.

Nonetheless, I believe I've asked about another thing. I want to build a manifold from this quotient therefore I need open covering. I need to show that $p(O)$ is open for some $O\subset \mathbb{R}\times(-1,1)$.

 

[fresh_42]

I am slightly confused. What kind of $U$s are considered here? If open, then I completely fine with this since canonical projection is continuous by definition.

$$
\mathcal{T}=\{U\subseteq M\,|\,p^{-1}(U)\subseteq \mathbb{R}\times (-1,1)\text{ is open}\}
$$
is meant to be the definition of the quotient topology, the definition of open sets.

Nonetheless, I believe I've asked about another thing. I want to build a manifold from this quotient therefore I need open covering. I need to show that $p(O)$ is open for some $O\subset \mathbb{R}\times(-1,1)$.

Set $p(O)=:U.$ Then $p^{-1}(U) =\bigcup_{k\in \mathbb{Z}} \phi_k(O) \subseteq \mathbb{R}\times (-1,1)$ is as a union of open sets open. Thus $U\in \mathcal{T}$ what had to be shown.

 

[Korybut]

Can I put it this way?
If the inverse image of the set $U$ of the canonical projection to the quotient is open then $U$ is open.

 

[fresh_42]

Can I put it this way?
If the inverse image of the set $U$ of the canonical projection to the quotient is open then $U$ is open.

Yes. This is the definition of the quotient topology. It includes just all sets that are necessary to make the projection a continuous function.

 

[Korybut]

Yes. This is the definition of the quotient topology. It includes just all sets that are necessary to make the projection a continuous function.

Many thanks! Can you provide some easy readable reference on this? Currently it looks like a pure magic of definitions

 

[fresh_42]

Many thanks! Can you provide some easy readable reference on this? Currently it looks like a pure magic of definitions

It is in a way pure magic. The idea is the following:

Given a set $M$ equipped with a topology $\mathcal{T}_M$, i.e. a definition which sets are called open, together with some properties: $\emptyset , M\in \mathcal{T}_M$, finite intersections and arbitrary unions of sets in $\mathcal{T}_M$ must also be in $\mathcal{T}_M.$

If we consider those sets as topological spaces, then we speak of the category of topological spaces. The topological spaces are the objects, and the continuous functions are the mappings between topological spaces. Hence open sets and continuity are the essential properties here.

In order to be a useful tool (category), we want to have some standards. $(M,\mathcal{T}_M)$ should allow subspaces, intersections, and unions to be topological spaces on their own again. We achieve this by setting for a subset $A\subseteq M$
$$
\mathcal{T}_A := \{U\subseteq A\,|\,\exists O \in \mathcal{T}_M\, : \,U=O\cap A\}
$$
which is the induced topology on $A$, i.e. what we call open sets in the topology of $A$.

Your example with the Möbius strip shows that this isn't enough. We also want to have direct products of topological spaces to be a topological space again, and quotient spaces of topological spaces to be a topological space again. Now, remember the functions between such spaces! We want them to be continuous. So we choose the smallest possible collection of sets as open sets, i.e. as topology, such that the projections on a direct factor, the inclusions of a factor, and the projection along a quotient are all continuous. This way $\mathbb{R}\times (-1,1)$ and $\mathbb{R}\times (-1,1)/\sim$ are again topological spaces and the natural functions that come with them will automatically be continuous.

 

[fresh_42]

Can you provide some easy readable reference on this?

Not really, will say not in English. I could only recommend sources in German, and I just found a source by one of the founding fathers himself. However, Urysohn wrote in French. There are probably also very good sources in Russian, considering the history of topology, but I do not know them. (Topology basically started as a German-Russian joint venture ;-))

A good method to get a quick supply is to google "lecture notes + topology". "Topologie" if you want German results, "Топология" if you are looking for Russian results.