- #1

Korybut

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- 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.

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.