# Thread: Decomposition space - Quotient spaces

1. ## Decomposition space - Quotient spaces

I want to prove that the decomposition space is the quotient topology.
But I have a lot of trouble understanding the definition 9.5 (to be able to solve Theorem 9.6). I don't really understand what F and F are.

Help appreciated!

Here's the definition & the theorem:

2. Originally Posted by sssitex
I want to prove that the decomposition space is the quotient topology.
But I have a lot of trouble understanding the definition 9.5 (to be able to solve Theorem 9.6). I don't really understand what F and F are.

Help appreciated!

Here's the definition & the theorem:
Here's the idea. Take $\displaystyle \mathbb{R}$ as a simple example. Then, we can form a decomposition as you call it (I'd call it a partition) by $\displaystyle \mathcal{D}=\left\{(z,z+1):z\in\mathbb{Z}\}\cup\le ft\{\{z\}:z\in\mathbb{Z}\right\}$ evidently $\displaystyle \displaystyle \mathbb{R}=\coprod_{D\in\mathcal{D}}D$. So, I want to topologize $\displaystyle \mathcal{D}$ by saying that $\displaystyle \mathcal{F}\subseteq\mathcal{D}$ is open if and only if $\displaystyle \displaystyle \bigcup_{F\in\mathcal{F}}F$ is open in $\displaystyle \mathbb{R}$. For example, $\displaystyle \left\{(z,z+1):z\in\mathbb{Z}\right\}=\mathcal{F}$ is open since $\displaystyle \displaystyle \bigcup_{F\in\mathcal{F}}F=\bigcup_{z\in\mathbb{Z} }(z,z+1)=\mathbb{R}-\mathbb{Z}$ which is open. Whereas $\displaystyle \displaystyle \mathcal{F}'=\left\{\{z\}:z\in\mathbb{Z}\right\}$ is not open since $\displaystyle \displaystyle \bigcup_{F\in\mathcal{F}'}F=\bigcup_{z\in\mathbb{Z }}\{z\}=\mathbb{Z}$ is not open in $\displaystyle \mathbb{R}$. Make sense?

For the actual exercise, if it makes it easier for you to think about it, define [tex]\sim [tex] on $\displaystyle \mathbb{R}$ by saying $\displaystyle x\sim y$ if and only if there exists $\displaystyle D\in\mathcal{D}$ such that $\displaystyle x,y\in D$. Then, the quotient map is $\displaystyle \pi:\mathbb{R}\to\mathcal{D}:x\mapsto[x]$ where $\displaystyle [x]$ is evidently the equivalence class of $\displaystyle x$ under $\displaystyle \sim$.