example of quotient topology

Koen just a moment. 1 answers, 0 views

if we work in $\mathbb{R}^2\(0,0)$ with euclidean topology and we set following equivalence relation $P$ on this space: $(x,y)P(x',y')$ iff there exists $a$ in $\mathbb{R}^2\(0,0)$ such that $(x,y) = a(x',y').$ how we define the open sets in quotient space $\mathbb{R}^2-\{(0,0)\}/P.$
Let $f: \mathbb{R}^2\(0,0) \to \mathbb{R}^2\(0,0)/P$ the quotient function: then the open sets in $\mathbb{R}^2\(0,0)/P$ are defined like: $\{f(A) \mid A\text{ open and saturated }\}$ but I have troubles by doing it for this example. Can someone help me?

1 Answers

Brian M. Scott 08/22/2013.

I’m going to assume that your $R$ is really $\Bbb R$, the set of real numbers.

HINT: Note that $\langle x,y\rangle\mathbin{P}\langle x',y'\rangle$ if and only if $\langle x,y\rangle$ and $\langle x',y'\rangle$ lie on the same straight line through the origin of $\Bbb R^2$. For $0\le\theta<\pi$ let $L_\theta'$ be the line through the origin containing the point $\langle\cos\theta,\sin\theta\rangle$, and let $L_\theta=L_\theta'\setminus\{\langle 0,0\rangle\}$. The sets $L_\theta$ for $\theta\in[0,\pi)$ are the $P$-equivalence classes and therefore also the fibres of the quotient map $f$ and the points of the quotient space. This means that you can identify the points of the quotient space with points of the interval $[0,\pi)$: each $L_\theta$ is a point of the quotient space, and it corresponds to the point $\theta$ in the set $[0,\pi)$.

To figure out the quotient topology $\tau$ on $[0,\pi)$, remember that a subset $U$ of $[0,\pi)$ is in $\tau$ if and only if $\bigcup_{\theta\in U}L_\theta$ is open in $\Bbb R^2\setminus\{\langle 0,0\rangle\}$. (It may help to imagine bending the interval $[0,\pi)$ around into a circle, to get the space obtained by identifying $0$ and $\pi$ in the closed interval $[0,\pi]$.)


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