# Topology Quotient Space

• May 17th 2008, 01:08 PM
TXGirl
Topology Quotient Space
In Topology 2ed by Munkres, the following definition is given of a quotient map:

Let $X,Y$ be topological spaces; let $p:{X}\rightarrow{Y}$ be a surjective map. The map $p$ is said to be a quotient map provided a subset ${U}\subset{Y}$ is open in $Y$ if and only if ${p^{-1}}(U)$ is open in $X$.

Then he gives the following example:

Let $X$ be the subspace $[0,1]\cup[2,3]$ of $\mathbb{R}$. The map $p:{X}\rightarrow{Y}$ defined by:

$p(x)=x$ if $x\in [0,1]$
$p(x)={(x-1)}$ if $x\in [2,3]$

In order to show that this map is a quotient map, one needs to show that is is surjective, continuous and closed. The first two are clear, however I am not so sure about how to verify that it is a closed map.

Also, he notes that the map is not open, saying as proof that the image of the "open set" $[0,1]$ of $X$ is not open in $Y$.

(i) how to verify that the map is a closed map
(ii) how $[0,1]$ can be considered open in $X$
(iii) how then is image of $[0,1]$ not open in $Y$
• May 17th 2008, 01:19 PM
TXGirl
I guess for (ii), I simply say that ${X-[0,1]}$ is $[2,3]$, which is clearly closed $\longrightarrow[0,1]$ is open in $X$, but what about (i) and (iii)?
• May 17th 2008, 01:32 PM
TXGirl
Sorry for the bother, I got it:

(i) $p([0,1])=[0,1]$ is closed in $Y$ and $p([2,3])=[1,2]$ is also closed in $Y$ thus $p:X \rightarrow Y$ is a closed map

(iii) $p([0,1])=[0,1]$ and $(Y-[0,1])=]1,2]$, which is not closed; thus, $p([0,1])$ is not open in $Y$, and so it follows that $p:X \rightarrow Y$ is not an open map
• May 17th 2008, 01:32 PM
Plato
What is Y? Is it $Y = [0,2]$? Is $[0,1]$ open in Y?
• May 17th 2008, 01:42 PM
TXGirl
Quote:

Originally Posted by Plato
What is Y? Is it $Y = [0,2]$? Is $[0,1]$ open in Y?

$X=[0,1]\cup[2,3] \subset \mathbb{R}$
$Y=[0,2] \subset \mathbb{R}$

$p$ is a map from $X$ to $Y$.

$[0,1]$ is open in $X$ because its complement in $X$ is $[2,3]$, which is closed.

$p([0,1])$ is not open in $Y$ because its complement in $Y$ is $]1,2]$, which is not closed.

The complement of $[0,1]$ in $Y$ is $]1,2]$, which is not closed, so no, $[0,1]$ is not open in $Y$