# 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.