# Math4202 Topology II (Lecture 2)

## Reviewing quotient map

Recall from last lecture example (Example 4 form Munkers):

A map of wrapping closed unit circle to $S^2$, where $f:\mathbb{R}^2\to S^2$ maps everything outside of circle to south pole $s$.

To show it is a quotient space, we need to show that $f$:

1. is continuous (every open set in $S^2$ has reverse image open in $\mathbb{R}^2$)
2. surjective (trivial)
3. with the property that $U\subset S^2$ is open if and only if $f^{-1}(U)$ is open in $\mathbb{R}^2$.

- If $A\subseteq S^2$ is open, then $f^{-1}(A)$ is open in $\mathbb{R}^2$. (consider the basis, the set of circle in $\mathbb{R}^2$, they are mapped to closed sets in $S^2$)

- If $f^{-1}(A)$ is open in $\mathbb{R}^2$, then $A$ is open in $S^2$.
  - If $s\notin A$, then $f$ is a bijection, and $A$ is open in $S^2$.
  - If $s\in A$, then $f^{-1}(A)$ is open and contains the complement of set $S=\{(x,y)|x^2+y^2\geq 1\}=f^{-1}(\{s\})$, therefore there exists $U=\bigcup_{x\in S} B_{\epsilon _x}(x)$ is open in $\mathbb{R}^2$, $U\subseteq f^{-1}(A)$, $f^{-1}(\{s\})\subseteq U$.
  - Since $\partial f^{-1}(\{s\})$ is compact, we can even choose $U$ to be the set of the following form
  - $\{(x,y)|x^2+y^2>1-\epsilon\}$ for some $1>\epsilon>0$.
  - So $f(U)$ is an open set in $A$ and contains $s$.
  - $s$ is an interior point of $A$.
  - Other oint $y$ in $A$ follows the arguments in the first case.

### Quotient space

#### Definition of quotient topology induced by quotient map

If $X$ is a topological space and $A$ is a set and if $p:X\to A$ is surjective, there exists exactly one topology $\mathcal{T}$ on $A$ relative to which $p$ is a quotient map.

$$
\mathcal{T} \coloneqq \{U|f^{-1}(U)\text{ is open in }X\}
$$

and $\mathcal{T}$ is called the quotient topology on $A$ induced by $p$.

#### Definition of quotient topology induced by equivalence relation

Let $X$ be a topological space, and let $X^*$ be a partition of $X$ into disjoint subsets whose union is $X$. Let $p:X\to X^*$ be the surjective map that sends each $x\in X$ to the unique $A\in X^*$ such that each point of $X$ to the subset containing the point. In the quotient topology induced by $p$, the space $X^*$ is called the associated quotient space.

<details>
<summary>Example of quotient topology induced by equivalence relation</summary>

Consider $S^n$  and $x\sim -x$, then the induced quotient topology is $\mathbb{R}P^n$ (the set of lines in $\mathbb{R}^n$ passing through the origin).

</details>

#### Theorem about a quotient map and quotient topology

Let $p:X\to Y$ be a quotient map; and $A$ be a subspace of $X$, that is **saturated** with respect to $p$: Let $q:A\to p(A)$ be the restriction of $p$ to $A$.

1. If $A$ is either open or closed in $X$, then $q$ is a quotient map.
2. If $p$ is either open or closed, then $q$ is a quotient map.

> [!NOTE]
>
> Recall the definition of saturated set:
>
> $\forall y\in Y$, consider the set $f^{-1}(\{y\})\subset X$, if $f^{-1}(\{y\})\cap A\neq \emptyset$, then $f^{-1}(\{y\})\subseteq A$. _sounds like connectedness_
>
> That is equivalent to say that $A$ is a union of $f^{-1}(\{y\})$ for some $y\in Y$.