1. ## Topology Quotient Space

In Topology 2ed by Munkres, the following definition is given of a quotient map:

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

Then he gives the following example:

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

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

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" $\displaystyle [0,1]$ of $\displaystyle X$ is not open in $\displaystyle Y$.

(i) how to verify that the map is a closed map
(ii) how $\displaystyle [0,1]$ can be considered open in $\displaystyle X$
(iii) how then is image of $\displaystyle [0,1]$ not open in $\displaystyle Y$

2. I guess for (ii), I simply say that $\displaystyle {X-[0,1]}$ is $\displaystyle [2,3]$, which is clearly closed $\displaystyle \longrightarrow[0,1]$ is open in $\displaystyle X$, but what about (i) and (iii)?

3. Sorry for the bother, I got it:

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

(iii) $\displaystyle p([0,1])=[0,1]$ and $\displaystyle (Y-[0,1])=]1,2]$, which is not closed; thus, $\displaystyle p([0,1])$ is not open in $\displaystyle Y$, and so it follows that $\displaystyle p:X \rightarrow Y$ is not an open map

4. What is Y? Is it $\displaystyle Y = [0,2]$? Is $\displaystyle [0,1]$ open in Y?

5. Originally Posted by Plato
What is Y? Is it $\displaystyle Y = [0,2]$? Is $\displaystyle [0,1]$ open in Y?
$\displaystyle X=[0,1]\cup[2,3] \subset \mathbb{R}$
$\displaystyle Y=[0,2] \subset \mathbb{R}$

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

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

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

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