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content/Math4201/Math4201_L18.md
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@@ -110,79 +110,6 @@ 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 ca
#### Proposition for continuous and quotient maps
Let $X,Y,Z$ be topological spaces. $p$ is a quotient map from $X$ to $Y$ and $g$ is a continuous map from $X$ to $Z$.
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 $g=\hat{g}\circ f$.
Moreover, if for any $y\in Y$, the map $g$ is constant on $p^{-1}(y)$, then there is a continuous map $f: Y\to Z$ satisfying $f\circ p=g$.
<details>
<summary>Proof</summary>
For any $y\in Y$, take $x\in X$ such that $p(x)=y$ (since $p$ is surjective).
Define $f(y)\coloneqq g(x)$.
Note that this is well-defined and it doesn't depend on the specific choice of $x$ that $p(x)=y$ because $g$ is constant on $p^{-1}(y)$.
Then we check that $f$ is continuous.
Let $U\subseteq Z$ be open. Then we want to show that $f^{-1}(U)\subseteq Y$ is open.
Since $p$ is a quotient map, this is equivalent to showing that $p^{-1}(f^{-1}(U))\subseteq X$ is open. Note that $p^{-1}(f^{-1}(U))=g^{-1}(U)$.
Since $g$ is continuous, $g^{-1}(U)$ is open in $X$.
Since $g^{-1}(U)$ is open in $X$, $p^{-1}(g^{-1}(U))$ is open in $Y$.
</details>
> In general, $p^{-1}(y)$ is called the **fiber** of $p$ over $y$. The $g$ must be constant on the fiber.
>
> We may define $p^{-1}(y)$ as the equivalence class of $y$ if $p$ is defined using the equivalence relation. By definition $p^{-1}([x])$ is the element of $x$ that are $\sim x$.
#### Additional to the proposition
Note that $f$ is unique.
It is not hard to see that $f$ is a quotient map if and only if $g$ is a quotient map. (check book for detailed proofs)
#### Definition of saturated map
Let $p:X\to Y$ be a quotient map. We say $A\subseteq X$ is **saturated** by $p$ if $A=p^{-1}(B)$ for some $B\subseteq Y$.
Equivalently, if $x\in A$, then $p^{-1}(p(x))\subseteq A$.
#### Proposition for quotient maps from saturated sets
Let $p:X\to Y$ be a quotient map and $q$ be given by restriction of $p$ to $A\subseteq X$. $q:A\to p(A)$, $q(x)=p(x),x\in A$.
Assume that $A$ is saturated by $p$.
1. If $A$ is closed or open, then $q$ is a quotient map.
2. If $p$ is closed or open, then $q$ is a quotient map.
<details>
<summary>Proof</summary>
We prove 1 and assume that $A$ is open, (the closed case is similar).
clearly, $q:A\to p(A)$ is surjective.
In general, restricting the domain and the range of a continuous map is continuous.
Since $A$ is saturated by $p$, then $p^{-1}(p(A))=A$ is open, so $p(A)$ is open because $p$ is a quotient map. Let $V\subseteq p(A)$ and $q^{-1}(V)\subseteq A$ is open. Then $q^{-1}(V)=p^{-1}(V)$.
(i) $q^{-1}(V)\subseteq p^{-1}(V)$: $x\in q^{-1}(V)\implies q(x)\in V$. Then $p(x)=q(x)\in V$
(ii) $p^{-1}(V)\subseteq q^{-1}(V)$: $x\in p^{-1}(V)\implies p(x)\in V\subseteq p(A)$. This implies that $x\in p^{-1}(p(A))=A$ since $A$ is saturated by $p$. Therefore $x\in q^{-1}(V)$.
Since $A$ is open in $X$, any open subspace of $A$ is open in $X$. In particular, $q^{-1}(V)=p^{-1}(V)$ is open in $X$.
Since $p$ is a quotient map, and $p^{-1}(V)$ is open in $X$, $V$ is open in $Y$. So $V\subseteq p(A)$ is open in $Y$.
This shows $q$ is a quotient map.
---
We prove 2 next time...
</details>
Continue next week.

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## Quotient topology
### More propositions
#### Proposition for continuous and quotient maps
Let $X,Y,Z$ be topological spaces. $p$ is a quotient map from $X$ to $Y$ and $g$ is a continuous map from $X$ to $Z$.
Moreover, if for any $y\in Y$, the map $g$ is constant on $p^{-1}(y)$, then there is a continuous map $f: Y\to Z$ satisfying $f\circ p=g$.
<details>
<summary>Proof</summary>
For any $y\in Y$, take $x\in X$ such that $p(x)=y$ (since $p$ is surjective).
Define $f(y)\coloneqq g(x)$.
Note that this is well-defined and it doesn't depend on the specific choice of $x$ that $p(x)=y$ because $g$ is constant on $p^{-1}(y)$.
Then we check that $f$ is continuous.
Let $U\subseteq Z$ be open. Then we want to show that $f^{-1}(U)\subseteq Y$ is open.
Since $p$ is a quotient map, this is equivalent to showing that $p^{-1}(f^{-1}(U))\subseteq X$ is open. Note that $p^{-1}(f^{-1}(U))=g^{-1}(U)$.
Since $g$ is continuous, $g^{-1}(U)$ is open in $X$.
Since $g^{-1}(U)$ is open in $X$, $p^{-1}(g^{-1}(U))$ is open in $Y$.
</details>
> In general, $p^{-1}(y)$ is called the **fiber** of $p$ over $y$. The $g$ must be constant on the fiber.
>
> We may define $p^{-1}(y)$ as the equivalence class of $y$ if $p$ is defined using the equivalence relation. By definition $p^{-1}([x])$ is the element of $x$ that are $\sim x$.
#### Additional to the proposition
Note that $f$ is unique.
It is not hard to see that $f$ is a quotient map if and only if $g$ is a quotient map. (check book for detailed proofs)
#### Definition of saturated map
Let $p:X\to Y$ be a quotient map. We say $A\subseteq X$ is **saturated** by $p$ if $A=p^{-1}(B)$ for some $B\subseteq Y$.
Equivalently, if $x\in A$, then $p^{-1}(p(x))\subseteq A$.
#### Proposition for quotient maps from saturated sets
Let $p:X\to Y$ be a quotient map and $q$ be given by restriction of $p$ to $A\subseteq X$. $q:A\to p(A)$, $q(x)=p(x),x\in A$.
Assume that $A$ is saturated by $p$.
1. If $A$ is closed or open, then $q$ is a quotient map.
2. If $p$ is closed or open, then $q$ is a quotient map.
<details>
<summary>Proof</summary>
We prove 1 and assume that $A$ is open, (the closed case is similar).
clearly, $q:A\to p(A)$ is surjective.
In general, restricting the domain and the range of a continuous map is continuous.
Since $A$ is saturated by $p$, then $p^{-1}(p(A))=A$ is open, so $p(A)$ is open because $p$ is a quotient map. Let $V\subseteq p(A)$ and $q^{-1}(V)\subseteq A$ is open. Then $q^{-1}(V)=p^{-1}(V)$.
(i) $q^{-1}(V)\subseteq p^{-1}(V)$: $x\in q^{-1}(V)\implies q(x)\in V$. Then $p(x)=q(x)\in V$
(ii) $p^{-1}(V)\subseteq q^{-1}(V)$: $x\in p^{-1}(V)\implies p(x)\in V\subseteq p(A)$. This implies that $x\in p^{-1}(p(A))=A$ since $A$ is saturated by $p$. Therefore $x\in q^{-1}(V)$.
Since $A$ is open in $X$, any open subspace of $A$ is open in $X$. In particular, $q^{-1}(V)=p^{-1}(V)$ is open in $X$.
Since $p$ is a quotient map, and $p^{-1}(V)$ is open in $X$, $V$ is open in $Y$. So $V\subseteq p(A)$ is open in $Y$.
This shows $q$ is a quotient map.
---
We prove 2 next time...
</details>

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# 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.

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@@ -23,4 +23,5 @@ export default {
Math4201_L17: "Topology I (Lecture 17)",
Math4201_L18: "Topology I (Lecture 18)",
Math4201_L19: "Topology I (Lecture 19)",
Math4201_L20: "Topology I (Lecture 20)",
}