diff --git a/content/Math4201/Math4201_L18.md b/content/Math4201/Math4201_L18.md
index 4d839ec..1b71259 100644
--- a/content/Math4201/Math4201_L18.md
+++ b/content/Math4201/Math4201_L18.md
@@ -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 $f\circ p=g$.
-
-
-Proof
-
-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 $Y$.
-
-
-
-> 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.
-
-
-Proof
-
-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...
-
-
\ No newline at end of file
+Continue next week.
\ No newline at end of file
diff --git a/content/Math4201/Math4201_L19.md b/content/Math4201/Math4201_L19.md
index fe7e4ab..ae49ca6 100644
--- a/content/Math4201/Math4201_L19.md
+++ b/content/Math4201/Math4201_L19.md
@@ -2,3 +2,83 @@
## 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 $f\circ p=g$.
+
+
+Proof
+
+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 $Y$.
+
+
+
+> 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.
+
+
+Proof
+
+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...
+
+
\ No newline at end of file
diff --git a/content/Math4201/Math4201_L20.md b/content/Math4201/Math4201_L20.md
new file mode 100644
index 0000000..c6607bd
--- /dev/null
+++ b/content/Math4201/Math4201_L20.md
@@ -0,0 +1,121 @@
+# 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).
+
+
+Proof
+
+$q$ 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 $p(A)$.
+
+
+
+### 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.
+
+
+Example of simplicial complex
+
+$n=0$,
+
+$\Delta^0=\{v_0\}$
+
+$n=1$,
+
+$\Delta^1=\{t_0v_0+t_1v_1\vert t_0+t_1=1\}$, 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 $v_0,v_1,v_2$.
+
+
+
+> [!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 $[v_{i_0},\dots,v_{i_n}]$.
+
+
+Example of faces for simplicial complex
\
+
+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).
+
+
+
+#### 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 $v\in V$, $\{v\}\in K$.
+
+
+Example of abstract simplicial complex
+
+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.
+
+
+
+#### 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.
\ No newline at end of file
diff --git a/content/Math4201/_meta.js b/content/Math4201/_meta.js
index e87a945..4624b9b 100644
--- a/content/Math4201/_meta.js
+++ b/content/Math4201/_meta.js
@@ -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)",
}