115 lines
3.7 KiB
Markdown
115 lines
3.7 KiB
Markdown
# Math4201 Topology I (Lecture 18)
|
|
|
|
## Quotient topology
|
|
|
|
Let $(X,\mathcal{T})$ be a topological space and $X^*$ be a set, $q:X\to X^*$ is a surjective map. The quotient topology on $X^*$:
|
|
|
|
$U\subseteq X^*$ is open $\iff q^{-1}(U)$ is open in $X$.
|
|
|
|
Equivalently,
|
|
|
|
$Z\subseteq X^*$ is closed $\iff q^{-1}(Z)$ is closed in $X$.
|
|
|
|
### Open maps
|
|
|
|
Let $(X,\mathcal{T})$ and $(Y,\mathcal{T}')$ be two topological spaces
|
|
|
|
Let $f:X\to Y$ is a quotient map if and only if $f$ is surjective and
|
|
|
|
$U\subseteq Y$ is open $\iff f^{-1}(U)$ is open
|
|
|
|
or equivalently
|
|
|
|
$Z\subseteq Y$ is closed $\iff f^{-1}(Z)$ is closed.
|
|
|
|
#### Definition of open map
|
|
|
|
Let $X\to Y$ be **continuous**. We say $f$ is open if for any $V\subseteq X$ be open, $f(V)$ is open in $Y$.
|
|
|
|
Let $X\to Y$ be **continuous**. We say $f$ is closed if for any $V\subseteq X$ be closed, $f(V)$ is closed in $Y$.
|
|
|
|
$$
|
|
ff^{-1}(U)=U\text{ if }f \text{ is surjective}=U\cap f(X)
|
|
$$
|
|
|
|
<details>
|
|
<summary>Examples of open maps</summary>
|
|
|
|
Let $X,Y$ be topological spaces. Define the projection map $\pi_X:X\times Y\to X$, $\pi_X(x,y)=x$.
|
|
|
|
This is a surjective continuous map $(Y\neq \phi)$
|
|
|
|
This map is open. If $U\subseteq X$ is open and $V\subseteq Y$ is open, then $U\times V$ is open in $X\times Y$ and such open sets form a basis.
|
|
|
|
$\pi_X(U\times V)=\begin{cases}
|
|
U&\text{ if }V\neq \emptyset\\
|
|
\emptyset &\text{ if }V=\emptyset
|
|
\end{cases}$
|
|
|
|
In particular, image of any such open set is open. Since any open $W\subseteq X\times Y$ is a union of such open sets.
|
|
|
|
$W=\bigcup_{\alpha\in I}U_\alpha\times V\alpha$
|
|
|
|
$\pi_X(W)=\pi_X(\bigcup_{\alpha\in I}U_\alpha\times V_\alpha)=\bigcup_{\alpha\in I}\pi_X(U_\alpha\times V_\alpha)=\bigcup_{\alpha\in I}U_\alpha$
|
|
|
|
is open in $X$.
|
|
|
|
However, $\pi_X$ is not necessarily a closed map.
|
|
|
|
Let $X=Y=\mathbb{R}$ and $X\times Y=\mathbb{R}^2$
|
|
|
|
$Z\subseteq \mathbb{R}^2=\{(x,y)\in\mathbb{R}^2|x\neq 0, y=\frac{1}{x}\}$ is a closed set in $\mathbb{R}^2$
|
|
|
|
$\pi_X(Z)=\mathbb{R}\setminus \{0\}$ is not closed.
|
|
|
|
---
|
|
|
|
Let $X=[0,1]\cup [2,3]$, $Y=[0,2]$ with subspace topology on $\mathbb{R}$
|
|
|
|
Let $f:X\to Y$ be defined as:
|
|
|
|
$$
|
|
f(x)=\begin{cases}
|
|
x& \text{ if } x\in [0,1]\\
|
|
x-1& \text{ if }x\in [2,3]
|
|
\end{cases}
|
|
$$
|
|
|
|
$f$ is continuous and surjective, $f$ is closed $Z\subseteq [0,1]\cup [2,3]=Z_1\cup Z_2$, $Z_1\subseteq [0,1],Z_2\subseteq [2,3]$ is closed, $f(Z)=f(Z_1)\cup f(Z_2)$ is closed in $X$.
|
|
|
|
But $f$ is not open. Take $U=[0,1]\subseteq X$, $f=[0,1]\subseteq [0,2]$ is not open because of the point $1$.
|
|
|
|
> In general, and closed surjective map is a quotient map. In particular, this is an example of a closed surjective quotient map which is not open.
|
|
|
|
</details>
|
|
|
|
Let $f$ be a surjective open map. Then $f$ is a quotient map:
|
|
|
|
$U\subseteq Y$ is open and $f$ is continuous, $\implies f^{-1}(U)\subseteq X$ is open
|
|
|
|
$f^{-1}(U)\subseteq X$ is open and $f$ is surjective and open, $\implies f(f^{-1}(U))=U$ is open.
|
|
|
|
#### Proposition of continuous and open maps
|
|
|
|
If $f$ is a continuous bijection, then $f$ is open. if and only if $f^{-1}$ is continuous.
|
|
|
|
<details>
|
|
<summary>Proof</summary>
|
|
|
|
To show $f^{-1}$ is continuous, we have to show for $U\subseteq X$ open. $(f^{-1})^{-1}(U)=f(U)\subseteq Y$ is open.
|
|
|
|
This is the same thing as saying that $f$ is open.
|
|
|
|
</details>
|
|
|
|
Let $f$ be a quotient map $f: X \to Y$, and $g$ be a continuous map $g:X\to Z$.
|
|
|
|
We want to find $\hat{g}$ such that $g=\hat{g}\circ f$.
|
|
|
|
If $x_1,x_2\in X$, such that $f(x_1)=f(x_2)$ and $g(x_1)\neq g(x_2)$, then we cannot find $\hat{g}$.
|
|
|
|
#### Proposition for continuous and quotient maps
|
|
|
|
Let $f$ and $g$ be as above. Moreover, for any $y\in Y$, all the points in $f^{-1}(y)$ are mapped to a single point by $g$. Then there is a unique continuous map $\hat{g}$ such that $g=\hat{g}\circ f$.
|
|
|
|
Continue next week. |