updates

content/Math4202/Exam_reviews/Math4202_E1.md

@@ -163,9 +163,9 @@ Define the product $f*g$ of $f$ and $g$ to be the map $h:[0,1]\to X$.
 
 #### Definition for equivalent classes of paths
 
-$\Pi_1(X,x)$ is the equivalent classes of paths starting and ending at $x$.
+$pi_1(X,x)$ is the equivalent classes of paths starting and ending at $x$.
 
-On $\Pi_1(X,x)$,, we define $\forall [f],[g],[f]*[g]=[f*g]$.
+On $pi_1(X,x)$,, we define $\forall [f],[g],[f]*[g]=[f*g]$.
 
 $$
 [f]\coloneqq \{f_i:[0,1]\to X|f_0(0)=f(0),f_i(1)=f(1)\}

content/Math4202/Math4202_L10.md

@@ -95,5 +95,5 @@ $\bar{f}_t=\bar{f}(1-ts)$ $s\in[\frac{1}{2},1]$.
 The fundamental group of $X$ at $x$ is defined to be
 
 $$
-(\Pi_1(X,x),*)
+(pi_1(X,x),*)
 $$

content/Math4202/Math4202_L11.md

@@ -8,9 +8,9 @@ The $*$ operation has the following properties:
 
 #### Properties for the path product operation
 
-Let $[f],[g]\in \Pi_1(X)$, for $[f]\in \Pi_1(X)$, let $s:\Pi_1(X)\to X, [f]\mapsto f(0)$ and $t:\Pi_1(X)\to X, [f]\mapsto f(1)$.
+Let $[f],[g]\in pi_1(X)$, for $[f]\in pi_1(X)$, let $s:pi_1(X)\to X, [f]\mapsto f(0)$ and $t:pi_1(X)\to X, [f]\mapsto f(1)$.
 
-Note that $t([f])=s([g])$, $[f]*[g]=[f*g]\in \Pi_1(X)$.
+Note that $t([f])=s([g])$, $[f]*[g]=[f*g]\in pi_1(X)$.
 
 This also satisfies the associativity. $([f]*[g])*[h]=[f]*([g]*[h])$.
 
@@ -51,33 +51,33 @@ Let $x_0\in X$. A path starting and ending at $x_0$ is called a loop based at $x
 The fundamental group of $X$ at $x$ is defined to be
 
 $$
-(\Pi_1(X,x),*)
+(pi_1(X,x),*)
 $$
 
-where $*$ is the product operation, and $\Pi_1(X,x)$ is the set o homotopy classes of loops in $X$ based at $x$.
+where $*$ is the product operation, and $pi_1(X,x)$ is the set o homotopy classes of loops in $X$ based at $x$.
 
 <details>
 <summary>Example of fundamental group</summary>
 
 Consider $X=[0,1]$, with subspace topology from standard topology in $\mathbb{R}$.
 
-$\Pi_1(X,0)=\{e\}$, (constant function at $0$) since we can build homotopy for all loops based at $0$ as follows $H(s,t)=(1-t)f(s)+t$.
+$pi_1(X,0)=\{e\}$, (constant function at $0$) since we can build homotopy for all loops based at $0$ as follows $H(s,t)=(1-t)f(s)+t$.
 
-And $\Pi_1(X,1)=\{e\}$, (constant function at $1$.)
+And $pi_1(X,1)=\{e\}$, (constant function at $1$.)
 
 ---
 
 Let $X=\{1,2\}$ with discrete topology.
 
-$\Pi_1(X,1)=\{e\}$, (constant function at $1$.)
+$pi_1(X,1)=\{e\}$, (constant function at $1$.)
 
-$\Pi_1(X,2)=\{e\}$, (constant function at $2$.)
+$pi_1(X,2)=\{e\}$, (constant function at $2$.)
 
 ---
 
 Let $X=S^1$ be the circle.
 
-$\Pi_1(X,1)=\mathbb{Z}$ (related to winding numbers, prove next week).
+$pi_1(X,1)=\mathbb{Z}$ (related to winding numbers, prove next week).
 
 </details>
 
@@ -85,7 +85,7 @@ A natural question is, will the fundamental group depends on the base point $x$?
 
 #### Definition for $\hat{\alpha}$
 
-Let $\alpha$ be a path in $X$ from $x_0$ to $x_1$. $\alpha:[0,1]\to X$ such that $\alpha(0)=x_0$ and $\alpha(1)=x_1$. Define $\hat{\alpha}:\Pi_1(X,x_0)\to \Pi_1(X,x_1)$ as follows:
+Let $\alpha$ be a path in $X$ from $x_0$ to $x_1$. $\alpha:[0,1]\to X$ such that $\alpha(0)=x_0$ and $\alpha(1)=x_1$. Define $\hat{\alpha}:pi_1(X,x_0)\to pi_1(X,x_1)$ as follows:
 
 $$
 \hat{\alpha}(\beta)=[\bar{\alpha}]*[f]*[\alpha]
@@ -93,12 +93,12 @@ $$
 
 #### $\hat{\alpha}$ is a group homomorphism
 
-$\hat{\alpha}$ is a group homomorphism between $(\Pi_1(X,x_0),*)$ and $(\Pi_1(X,x_1),*)$
+$\hat{\alpha}$ is a group homomorphism between $(pi_1(X,x_0),*)$ and $(pi_1(X,x_1),*)$
 
 <details>
 <summary>Proof</summary>
 
-Let $f,g\in \Pi_1(X,x_0)$, then $\hat{\alpha}(f*g)=\hat{\alpha}(f)\hat{\alpha}(g)$
+Let $f,g\in pi_1(X,x_0)$, then $\hat{\alpha}(f*g)=\hat{\alpha}(f)\hat{\alpha}(g)$
 
 $$
 \begin{aligned}
@@ -129,4 +129,4 @@ The other case is the same
 
 #### Corollary of fundamental group
 
-If $X$ is path-connected and $x_0,x_1\in X$, then $\Pi_1(X,x_0)$ is isomorphic to $\Pi_1(X,x_1)$.
+If $X$ is path-connected and $x_0,x_1\in X$, then $pi_1(X,x_0)$ is isomorphic to $pi_1(X,x_1)$.

content/Math4202/Math4202_L12.md

@@ -4,18 +4,18 @@
 
 ### Fundamental group
 
-Recall from last lecture, the $(\Pi_1(X,x_0),*)$ is a group, and for any two points $x_0,x_1\in X$, the group $(\Pi_1(X,x_0),*)$ is isomorphic to $(\Pi_1(X,x_1),*)$ if $x_0,x_1$ is path connected.
+Recall from last lecture, the $(pi_1(X,x_0),*)$ is a group, and for any two points $x_0,x_1\in X$, the group $(pi_1(X,x_0),*)$ is isomorphic to $(pi_1(X,x_1),*)$ if $x_0,x_1$ is path connected.
 
 > [!TIP]
 > 
-> How does the $\hat{\alpha}$ (isomorphism between $(\Pi_1(X,x_0),*)$ and $(\Pi_1(X,x_1),*)$) depend on the choice of $\alpha$ (path) we choose?
+> How does the $\hat{\alpha}$ (isomorphism between $(pi_1(X,x_0),*)$ and $(pi_1(X,x_1),*)$) depend on the choice of $\alpha$ (path) we choose?
 
 #### Definition of simply connected
 
 A space $X$ is simply connected if 
 
 - $X$ is [path-connected](https://notenextra.trance-0.com/Math4201/Math4201_L23/#definition-of-path-connected-space) ($\forall x_0,x_1\in X$, there exists a continuous function $\alpha:[0,1]\to X$ such that $\alpha(0)=x_0$ and $\alpha(1)=x_1$)
-- $\Pi_1(X,x_0)$ is the trivial group for some $x_0\in X$
+- $pi_1(X,x_0)$ is the trivial group for some $x_0\in X$
 
 <details>
 <summary>Example of simply connected space</summary>
@@ -59,7 +59,7 @@ $$
 
 #### Definition of group homomorphism induced by continuous map
 
-Let $h:(X,x_0)\to (Y,y_0)$ be a continuous map, define $h_*:\Pi_1(X,x_0)\to \Pi_1(Y,y_0)$ where $h(x_0)=y_0$. by $h_*([f])=[h\circ f]$.
+Let $h:(X,x_0)\to (Y,y_0)$ be a continuous map, define $h_*:pi_1(X,x_0)\to pi_1(Y,y_0)$ where $h(x_0)=y_0$. by $h_*([f])=[h\circ f]$.
 
 $h_*$ is called the group homomorphism induced by $h$ relative to $x_0$.
 
@@ -80,7 +80,7 @@ $$
 
 #### Theorem composite of group homomorphism
 
-If $h:(X,x_0)\to (Y,y_0)$ and $k:(Y,y_0)\to (Z,z_0)$ are continuous maps, then $k_* \circ h_*:\Pi_1(X,x_0)\to \Pi_1(Z,z_0)$ where $h_*:\Pi_1(X,x_0)\to \Pi_1(Y,y_0)$, $k_*:\Pi_1(Y,y_0)\to \Pi_1(Z,z_0)$,is a group homomorphism.
+If $h:(X,x_0)\to (Y,y_0)$ and $k:(Y,y_0)\to (Z,z_0)$ are continuous maps, then $k_* \circ h_*:pi_1(X,x_0)\to pi_1(Z,z_0)$ where $h_*:pi_1(X,x_0)\to pi_1(Y,y_0)$, $k_*:pi_1(Y,y_0)\to pi_1(Z,z_0)$,is a group homomorphism.
 
 <details>
 <summary>Proof</summary>
@@ -100,7 +100,7 @@ $$
 
 #### Corollary of composite of group homomorphism
 
-Let $\operatorname{id}:(X,x_0)\to (X,x_0)$ be the identity map. This induces $(\operatorname{id})_*:\Pi_1(X,x_0)\to \Pi_1(X,x_0)$.
+Let $\operatorname{id}:(X,x_0)\to (X,x_0)$ be the identity map. This induces $(\operatorname{id})_*:pi_1(X,x_0)\to pi_1(X,x_0)$.
 
 If $h$ is a homeomorphism with the inverse $k$, with
 
@@ -108,7 +108,7 @@ $$
 k_*\circ h_*=(k\circ h)_*=(\operatorname{id})_*=I=(\operatorname{id})_*=(h\circ k)_*
 $$
 
-This induced $h_*: \Pi_1(X,x_0)\to \Pi_1(Y,y_0)$ is an isomorphism.
+This induced $h_*: pi_1(X,x_0)\to pi_1(Y,y_0)$ is an isomorphism.
 
 #### Corollary for homotopy and group homomorphism
 

content/Math4202/Math4202_L24.md

@@ -0,0 +1,78 @@
+# Math4202 Topology II (Lecture 24)
+
+## Algebraic Topology
+
+### Deformation Retracts and Homotopy Type
+
+Recall from last lecture, let $h,k:(X,x_0)\to (Y,y_0)$ be continuous maps. If there exists a homotopy of $h,y$ such that $H:X\times I\to Y$ that $H(x_0,t)=y_0$.
+
+Then $h_*=k_*:\pi_1(X,x_0)\to \pi_1(Y,y_0)$.
+
+We can prove this by showing that all the loop $f:I\to X$ based at $x_0$, $h_*([f])=k_*([f])$.
+
+That is $[h\circ f]=[k\circ f]$.
+
+This is a function $I\times I \to Y$ by $(s,t)\mapsto H(f(s),t)$.
+
+We need to show that this is a homotopy between $h\circ f$ and $k\circ f$.
+
+#### Theorem
+
+The Inclusion map $j:S^n\to \mathbb{R}^n-\{0\}$ induces on isomorphism of fundamental groups
+
+$$
+j_*:\pi_1(S^n)\to \pi_1(\mathbb{R}^n-\{0\})
+$$
+
+The function is injective.
+
+> Recall we showed that $S^1\to \mathbb{R}-\{0\}$ is injective by $x\mapsto \frac{x}{|x|}$.
+
+We want to show that $j_*\circ r_*=id_{\pi_1(S^n)}\quad r_*\circ j_*=id_{\pi_1(\mathbb{R}^n-\{0\})}$, then $r_*$, $j_*$ are isomorphism.
+
+<details>
+<summary>Proof</summary>
+
+**Homotopy is well defined**.
+
+Consider $H:(\mathbb{R}^n-\{0\})\times I\to \mathbb{R}^n-\{0\}$.
+
+Given $(x,t)\mapsto tx+(1-t)\frac{x}{\|x\|}$.
+
+Note that $(t-\frac{1-t}{\|x\|})x=0\implies t=0\land t=1$.
+
+So this map is well defined.
+
+**Base point is fixed**.
+
+On point $(1,0)$ (or anything on the sphere), $H(x,0)=x$.
+
+</details>
+
+#### Definition of deformation retract
+
+Let $A$ be a subspace of $X$, we say that $A$ is a deformation retract of $X$ if the identity map of $X$ is homotopic to a map that carries all $X$ to $A$ such that each point of $A$ remains fixed during the homotopy.
+
+Equivalently, there exists a homotopy $H:X\times I\to X$ such that:
+
+- $H(x,0)=x$ forall $x\in X$
+- $H(a,t)=a$ for all $a\in A$, $t\in I$
+- $H(x,1)\in A$ for all $x\in X$
+
+Equivalently,
+
+$r:H(x,1):X\to A$ is a retract.
+
+If we let $j:A\to X$ be the inclusion map, then $r\circ j=id_A$, and $j\circ r\sim id_X$ (with $A$ fixed.)
+
+<details>
+<summary>Example of deformation retract</summary>
+
+$S^1$ is a deformation retract of $\mathbb{R}^2-\{0\}$ 
+
+</details>
+
+#### Theorem for Deformation Retract
+
+If $A$ is a deformation retract of $X$, then $A$ and $X$ have the same fundamental group.
+

content/Math4202/Math4202_L9.md

@@ -7,7 +7,7 @@
 Consider the space of paths up to homotopy equivalence.
 
 $$
-\operatorname{Path}/\simeq_p(X) =\Pi_1(X)
+\operatorname{Path}/\simeq_p(X) =pi_1(X)
 $$
 
 We want to impose some group structure on $\operatorname{Path}/\simeq_p(X)$.
@@ -33,9 +33,9 @@ Define the product $f*g$ of $f$ and $g$ to be the map $h:[0,1]\to X$.
 
 #### Definition for equivalent classes of paths
 
-$\Pi_1(X,x)$ is the equivalent classes of paths starting and ending at $x$.
+$pi_1(X,x)$ is the equivalent classes of paths starting and ending at $x$.
 
-On $\Pi_1(X,x)$,, we define $\forall [f],[g],[f]*[g]=[f*g]$.
+On $pi_1(X,x)$,, we define $\forall [f],[g],[f]*[g]=[f*g]$.
 
 $$
 [f]\coloneqq \{f_i:[0,1]\to X|f_0(0)=f(0),f_i(1)=f(1)\}
@@ -141,5 +141,5 @@ Continue next time.
 The fundamental group of $X$ at $x$ is defined to be
 
 $$
-(\Pi_1(X,x),*)
+(pi_1(X,x),*)
 $$

content/Math4202/_meta.js

@@ -1,3 +1,5 @@
+import { MathJax } from "nextra/components";
+
 export default {
     index: "Course Description",
     "---":{
@@ -27,4 +29,5 @@ export default {
     Math4202_L21: "Topology II (Lecture 21)",
     Math4202_L22: "Topology II (Lecture 22)",
     Math4202_L23: "Topology II (Lecture 23)",
+    Math4202_L24: "Topology II (Lecture 24)",
 }

content/Math4302/Math4302_L25.md

@@ -0,0 +1,141 @@
+# Math4302 Modern Algebra (Lecture 25)
+
+Midterm next, next Wednesday
+
+## Rings
+
+### Definitions
+
+- commutative ring: elements $a\cdot b=b\cdot a$, $\forall a,b\in R$
+- ring with unity: elements $a\cdot 1=1\cdot a=a$, $\forall a\in R$
+- units: elements such that there is $a\cdot b=1$ for some $b\in R$.
+- division ring: every element $a\neq 0$ has a multiplicative inverse $a^{-1}$ such that $a\cdot a^{-1}=1$.
+- field: division ring that is commutative
+
+<details>
+<summary>Examples of division ring that is not a field</summary>
+
+Quaternions
+
+Let $i^2=-1$, $j^2=-1$, $k^2=-1$, with $ij=k$, $jk=i$, $ki=j$.
+
+$R=\{a+bi+ci+dj\mid a,b,c,d\in \mathbb{R}\}$
+
+$R$ is not commutative since $ij\neq ji$, but $R$ is a division ring.
+
+Let $x=a+bi+cj+dk$ be none zero, then $\bar{x}=a-bi-cj-dk$, $x^{-1}=\frac{\bar{x}}{a^2+b^2+c^2+d^2}$ is also non zero and $xx^{-1}=1$.
+
+</details>
+
+Recall from last time $\mathbb{Z}_n$ is a field if and only if $n$ is prime.
+
+#### Units in $\mathbb{Z}_n$ is coprime to $n$
+
+More generally, $[m]\in \mathbb{Z}_n$ is a unit if and only if $\operatorname{gcd}(m,n)=1$.
+
+<details>
+<summary>Proof</summary>
+
+Let $d=\operatorname{gcd}(m,n)$ and $[m]$ is a unit, then $\exists [x]\in \mathbb{Z}_n$ with $[m][z]=[1]$, so $mz\equiv 1\mod n$. so $mz-1=nt$ for some $t\in \mathbb{Z}$, but $d|m$, $d|t$, so $d|1$ implies $d=1$.
+
+If $d=1$, so $1=mr+ns$ for some $r,s\in \mathbb{Z}_n$. If $x=r\mod n$, then $[x]$ is the inverse of $[m]$. $mr\equiv 1\mod n\implies [m][x]=[1]$.
+
+</details>
+
+### Integral Domains
+
+#### Definition of zero divisors
+
+If $a,b\in R$ with $a,b\neq 0$ and $ab=0$, then $a,b$ are called zero divisors.
+
+<details>
+<summary>Example of zero divisors</summary>
+
+Consider $\mathbb{Z}_6$, then $2\cdot 3=0$, so $2$ and $3$ are zero divisors.
+
+And $4\cdot 3=0$, so $4$ and $3$ are zero divisors.
+
+> If $a$ is a unit, then $a$ is not a zero divisor.
+
+$ab=0\implies a^{-1}ab=0\implies 1b=0\implies b=0$.
+
+</details>
+
+> [!NOTE]
+>
+> If an element is not unit, it may not be a zero divisor.
+>
+> Consider $R=\mathbb{Z}$ and $2$ is not a unit, but $2$ is not a zero divisor.
+
+#### Zero divisors in $\mathbb{Z}_n$
+
+$[m]\in \mathbb{Z}_n$ is a zero divisor if and only if $\operatorname{gcd}(m,n)>1$ ($m$ is not a unit).
+
+<details>
+<summary>Proof</summary>
+
+If $d=\operatorname{gcd}(m,n)=1$, then $[m]$ is a unit, so $[m]$ is not a zero divisor.
+
+Therefore $[m]$ is a zero divisor if $\operatorname{gcd}(m,n)>1$.
+
+---
+
+If $d=\operatorname{gcd}(m,n)>1$, then $n=n_1d,m=m_1d, 1\leq n_1<n$.
+
+Then $mn_1=m_1dn_1=m_1n$, $n|mn_1$ $[m][n_1]=[0]$, $n_1\neq 0$, $[m]$ is a zero divisor.
+
+</details>
+
+#### Definition of integral domain
+
+A commutative ring with unity is called a integral domain (or just a domain) if it has no zero divisors.
+
+<details>
+<summary>Example of integral domain</summary>
+
+$\mathbb{Z}$ is a integral domain.
+
+---
+
+Any field is a integral domain.
+</details>
+
+#### Corollaries of integral domain
+
+If $R$ is a integral domain, then we have cancellation property $ab=ac,a\neq 0\implies b=c$.
+
+#### Units with multiplication forms a group
+
+If $R$ is a ring with unity, then the units in $R$ forms a group under multiplication.
+
+<details>
+<summary>Proof</summary>
+
+if $a,b$ are units, then $ab$ is a unit $(ab)^{-1}=b^{-1}a^{-1}$.
+
+</details>
+
+In particular, non-zero elements of any field form an abelian group under multiplication.
+
+<details>
+<summary>Example</summary>
+
+Consider $\mathbb{Z}_p$ field, then $(\{1,2,\cdots,p-1\},\cdot)$ forms an abelian group of size $p-1$.
+
+---
+
+Consider $\mathbb{Z}_5$, then we have a group of size $4$ under multiplication.
+
+- $1$ has order 1
+- $2$ has order 4 $2,4,3,1$.
+- $3$ has order 4 $3,4,2,1$.
+- $4$ has order 2 $4,1$.
+
+Therefore $\mathbb{Z}_5\simeq \mathbb{Z}_4$.
+
+---
+
+Therefore in $R=\mathbb{Z}_p$, $\mathbb{Z}_p^*=\{[1],[2],\cdots,[p-1]\}$ is a group of order $p-1$.
+
+Therefore, for every $a\in \mathbb{Z}_p$, $[a]^{p-1}=[1]$, then $a^{p-1}\equiv 1\mod p$ (Fermat's little theorem).
+</details>

content/Math4302/_meta.js

@@ -27,4 +27,5 @@ export default {
     Math4302_L22: "Modern Algebra (Lecture 22)",
     Math4302_L23: "Modern Algebra (Lecture 23)",
     Math4302_L24: "Modern Algebra (Lecture 24)",
+    Math4302_L25: "Modern Algebra (Lecture 25)",
 }