# Math4201 Topology I (Lecture 20)

## Quotient topology

### More propositions

#### Proposition for quotient maps in restrictions

Let $X,Y$ be topological spaces and $p:X\to Y$ is surjective and open/closed. Let $A\subseteq X$ be saturated by $p$, ($p^{-1}(p(A))=A$).

Then $q: A\to p(A)$ given by the restriction of $p$ is open/closed surjective map (In particular, it's a quotient map).

<details>
<summary>Proof</summary>

$q$ is surjective and continuous. Now assume $p$ is open and we will show that $q$ is also open. Any open subspace of $A$ is given as $U\cap A$ where $U$ is open in $X$. By definition, $q(U\cap A)=p(U\cap A)=p(U)\cap p(A)$

To see the second identity:

1. $p(U\cap A)\subseteq p(U)\cap p(A)$

$\forall y\in p(U\cap A)$, $y=p(x)$ with $x\in U\cap A$, since $x\in A$ and $x\in U$, $y=p(x)\in p(U)\cap p(A)$

2. $p(U)\cap p(A)\subseteq p(U\cap A)$

$\forall y\in p(U)\cap p(A)$, $y=p(x_1)$ with $x_1\in U$ and $y=p(x_2)$ with $x_2\in A$, since $x_1\in U$ and $x_2\in A$, $y=p(x_1)=p(x_2)\in p(U\cap A)$

So $x_1=x_2\in U\cap A$, $y=p(x_1)=p(x_2)\in p(U)\cap p(A)$, $y\in p(U\cap A)$.

Note that $p(U)\subseteq X$ is open by $p$ is an open map.

So $p(U)\cap p(A)$ is open in $p(A)$.

$q(U\cap A)=p(U\cap A)=p(U)\cap p(A)$ is open.

So $q$ is open in $p(A)$.

</details>

### Simplicial complexes (extra chapter)

#### Definition for simplicial complexes

Simplicial complexes are topological space with simplices ($n$ dimensional triangles) as their building blocks.

#### Definition for n dimensional simplex

Let $v_0,\dots,v_n$ be points in $\mathbb{R}^m$ such that $v_n-v_0$, $v_{n-1}-v_0$, $\cdots$, and $v_1-v_0$ are linearly independent in $\mathbb{R}^m$. (in particular $n\leq m$).

The $n$-dimensional simplex determined by $\{v_0,\dots,v_n\}$ is given as:

$$
\Delta^n\coloneqq [v_0,\dots,v_n]=\{t_0v_0+t_1v_1+\cdots+t_nv_n\vert t_i\geq 0, \sum_{i=0}^n t_i=1\}
$$

The coefficients $t_0,\dots,t_n$ are called barycentric coordinates.

<details>
<summary>Example of simplicial complex</summary>

$n=0$,

$\Delta^0=\{v_0\}$

$n=1$,

$\Delta^1=\{t_0v_0+t_1v_1\vert t_0+t_1=1\}$, this is the line segment between $v_0$ and $v_1$.

$n=2$,

$\Delta^2=\{t_0v_0+t_1v_1+t_2v_2\vert t_0+t_1+t_2=1\}$, this is the triangle with vertices $v_0,v_1,v_2$.

</details>

> [!NOTE]
>
> Every non-empty subset $\{v_{i_0},\dots,v_{i_k}\}$ of $\{v_0,\dots,v_n\}$ determines a $k$ dimensional simplex $[v_{i_0},\dots,v_{i_k}]\subseteq \Delta^n=[v_0,\dots,v_n]$. Inside the $n$ dimensional simplex $t_{i_0}v_{i_0}+\cdots+t_{i_n}v_{i_k}\in \Delta^n$. Where the coefficient $t_j$ of $v_j\notin \{v_{i_0},\dots,v_{i_n}\}$ is $0$.

Any such $k$ dimensional simplex is called a face of the simplex $[v_{i_0},\dots,v_{i_n}]$.

<details>
<summary>Example of faces for simplicial complex</summary>\

For a triangle $[v_0,v_1,v_2]$, the faces are $[v_0,v_1]$, $[v_0,v_2]$, and $[v_1,v_2]$ (the edges of the triangle).

</details>

#### Definition for abstract simplicial complex

Let $V$ be a finite or countable set, an abstract simplicial complex on $V$ is a collection of **finite non-empty subset** of $V$, denoted by $K$. And the two conditions are satisfied:

1. If $\sigma\in K$ and $\tau\subseteq \sigma$, then $\tau\in K$.

2. For any $v\in V$, $\{v\}\in K$.

<details>
<summary>Example of abstract simplicial complex</summary>

Let $V=\{a,b,c,d\}$.

If we want to include $\{a,b,c\}$, then we need to include $\{a,b\}$ and $\{b,c\}$, so we have $K=\{\{a,b,c\},\{a,b\},\{b,c\},\{a\},\{b\},\{c\},\{d\}\}$ is an abstract simplicial complex. 

</details>

#### Topological realization of abstract simplicial complex

Let $\bigsqcup_{\sigma\in K}\Delta^{|\sigma|-1}$ be the disjoint union of all $|\sigma|-1$ dimensional simplices in $K$.

$$
\tilde{X_k}=\bigsqcup_{\sigma\in K}\Delta^{|\sigma|-1}
$$

We use subspace topology to define a topology on $\Delta^n$ and the union of such topology for each $\Delta^{|\sigma|-1}$ defines a topology on $\tilde{X_k}$.

We define the equivalence relation $x\in \Delta_{\sigma}^{|\sigma|-1}\sim x'\in \Delta_{\sigma'}^{|\sigma'|-1}$ if $x\in \Delta_{\sigma'\cap \sigma}^{|\sigma'\cap \sigma|-1}\subseteq \Delta_{\sigma}^{|\sigma|-1}$. and $x'\in \Delta_{\sigma'\cap \sigma}^{|\sigma'\cap \sigma|-1}\subseteq \Delta_{\sigma'}^{|\sigma'|-1}$.

are the sample points of $\Delta_{\sigma\cap \sigma'}^{|\sigma\cap \sigma'|-1}$.

$X_K$ is the quotient space of $\tilde{X_k}$ by the equivalence relation.

Continue next time.