From 3f4479157bb7736ab746f38260f7e186bf636de2 Mon Sep 17 00:00:00 2001 From: Zheyuan Wu <60459821+Trance-0@users.noreply.github.com> Date: Fri, 20 Mar 2026 16:59:05 -0500 Subject: [PATCH] updates --- content/Math4202/Exam_reviews/Math4202_E1.md | 4 +- content/Math4202/Math4202_L10.md | 2 +- content/Math4202/Math4202_L11.md | 26 ++-- content/Math4202/Math4202_L12.md | 14 +- content/Math4202/Math4202_L24.md | 78 ++++++++++ content/Math4202/Math4202_L9.md | 8 +- content/Math4202/_meta.js | 3 + content/Math4302/Math4302_L25.md | 141 +++++++++++++++++++ content/Math4302/_meta.js | 1 + 9 files changed, 250 insertions(+), 27 deletions(-) create mode 100644 content/Math4202/Math4202_L24.md create mode 100644 content/Math4302/Math4302_L25.md diff --git a/content/Math4202/Exam_reviews/Math4202_E1.md b/content/Math4202/Exam_reviews/Math4202_E1.md index e9ef47f..bdce062 100644 --- a/content/Math4202/Exam_reviews/Math4202_E1.md +++ b/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 $\forall [f],[g],[f]*[g]=[f*g]$. $$ [f]\coloneqq \{f_i:[0,1]\to X|f_0(0)=f(0),f_i(1)=f(1)\} diff --git a/content/Math4202/Math4202_L10.md b/content/Math4202/Math4202_L10.md index 0475678..5034c82 100644 --- a/content/Math4202/Math4202_L10.md +++ b/content/Math4202/Math4202_L10.md @@ -95,5 +95,5 @@ $\bar{f}_t=\bar{f}(1-ts)$ $s\in[\frac{1}{2},1]$. The fundamental group of $X$ at $x$ is defined to be $$ -(\Pi_1(X,x),*) +(pi_1(X,x),*) $$ \ No newline at end of file diff --git a/content/Math4202/Math4202_L11.md b/content/Math4202/Math4202_L11.md index de0b896..0b79093 100644 --- a/content/Math4202/Math4202_L11.md +++ b/content/Math4202/Math4202_L11.md @@ -8,9 +8,9 @@ The $*$ operation has the following properties: #### Properties for the path product operation -Let $[f],[g]\in \Pi_1(X)$, for $[f]\in \Pi_1(X)$, let $s:\Pi_1(X)\to X, [f]\mapsto f(0)$ and $t:\Pi_1(X)\to X, [f]\mapsto f(1)$. +Let $[f],[g]\in pi_1(X)$, for $[f]\in pi_1(X)$, let $s:pi_1(X)\to X, [f]\mapsto f(0)$ and $t:pi_1(X)\to X, [f]\mapsto f(1)$. -Note that $t([f])=s([g])$, $[f]*[g]=[f*g]\in \Pi_1(X)$. +Note that $t([f])=s([g])$, $[f]*[g]=[f*g]\in pi_1(X)$. This also satisfies the associativity. $([f]*[g])*[h]=[f]*([g]*[h])$. @@ -51,33 +51,33 @@ Let $x_0\in X$. A path starting and ending at $x_0$ is called a loop based at $x The fundamental group of $X$ at $x$ is defined to be $$ -(\Pi_1(X,x),*) +(pi_1(X,x),*) $$ -where $*$ is the product operation, and $\Pi_1(X,x)$ is the set o homotopy classes of loops in $X$ based at $x$. +where $*$ is the product operation, and $pi_1(X,x)$ is the set o homotopy classes of loops in $X$ based at $x$.
Example of fundamental group Consider $X=[0,1]$, with subspace topology from standard topology in $\mathbb{R}$. -$\Pi_1(X,0)=\{e\}$, (constant function at $0$) since we can build homotopy for all loops based at $0$ as follows $H(s,t)=(1-t)f(s)+t$. +$pi_1(X,0)=\{e\}$, (constant function at $0$) since we can build homotopy for all loops based at $0$ as follows $H(s,t)=(1-t)f(s)+t$. -And $\Pi_1(X,1)=\{e\}$, (constant function at $1$.) +And $pi_1(X,1)=\{e\}$, (constant function at $1$.) --- Let $X=\{1,2\}$ with discrete topology. -$\Pi_1(X,1)=\{e\}$, (constant function at $1$.) +$pi_1(X,1)=\{e\}$, (constant function at $1$.) -$\Pi_1(X,2)=\{e\}$, (constant function at $2$.) +$pi_1(X,2)=\{e\}$, (constant function at $2$.) --- Let $X=S^1$ be the circle. -$\Pi_1(X,1)=\mathbb{Z}$ (related to winding numbers, prove next week). +$pi_1(X,1)=\mathbb{Z}$ (related to winding numbers, prove next week).
@@ -85,7 +85,7 @@ A natural question is, will the fundamental group depends on the base point $x$? #### Definition for $\hat{\alpha}$ -Let $\alpha$ be a path in $X$ from $x_0$ to $x_1$. $\alpha:[0,1]\to X$ such that $\alpha(0)=x_0$ and $\alpha(1)=x_1$. Define $\hat{\alpha}:\Pi_1(X,x_0)\to \Pi_1(X,x_1)$ as follows: +Let $\alpha$ be a path in $X$ from $x_0$ to $x_1$. $\alpha:[0,1]\to X$ such that $\alpha(0)=x_0$ and $\alpha(1)=x_1$. Define $\hat{\alpha}:pi_1(X,x_0)\to pi_1(X,x_1)$ as follows: $$ \hat{\alpha}(\beta)=[\bar{\alpha}]*[f]*[\alpha] @@ -93,12 +93,12 @@ $$ #### $\hat{\alpha}$ is a group homomorphism -$\hat{\alpha}$ is a group homomorphism between $(\Pi_1(X,x_0),*)$ and $(\Pi_1(X,x_1),*)$ +$\hat{\alpha}$ is a group homomorphism between $(pi_1(X,x_0),*)$ and $(pi_1(X,x_1),*)$
Proof -Let $f,g\in \Pi_1(X,x_0)$, then $\hat{\alpha}(f*g)=\hat{\alpha}(f)\hat{\alpha}(g)$ +Let $f,g\in pi_1(X,x_0)$, then $\hat{\alpha}(f*g)=\hat{\alpha}(f)\hat{\alpha}(g)$ $$ \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 $pi_1(X,x_1)$. diff --git a/content/Math4202/Math4202_L12.md b/content/Math4202/Math4202_L12.md index da09c26..75b19c1 100644 --- a/content/Math4202/Math4202_L12.md +++ b/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 $x_0\in X$
Example of simply connected space @@ -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.
Proof @@ -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)_* $$ -This induced $h_*: \Pi_1(X,x_0)\to \Pi_1(Y,y_0)$ is an isomorphism. +This induced $h_*: pi_1(X,x_0)\to pi_1(Y,y_0)$ is an isomorphism. #### Corollary for homotopy and group homomorphism diff --git a/content/Math4202/Math4202_L24.md b/content/Math4202/Math4202_L24.md new file mode 100644 index 0000000..1280e1c --- /dev/null +++ b/content/Math4202/Math4202_L24.md @@ -0,0 +1,78 @@ +# Math4202 Topology II (Lecture 24) + +## Algebraic Topology + +### Deformation Retracts and Homotopy Type + +Recall from last lecture, let $h,k:(X,x_0)\to (Y,y_0)$ be continuous maps. If there exists a homotopy of $h,y$ such that $H:X\times I\to Y$ that $H(x_0,t)=y_0$. + +Then $h_*=k_*:\pi_1(X,x_0)\to \pi_1(Y,y_0)$. + +We can prove this by showing that all the loop $f:I\to X$ based at $x_0$, $h_*([f])=k_*([f])$. + +That is $[h\circ f]=[k\circ f]$. + +This is a function $I\times I \to Y$ by $(s,t)\mapsto H(f(s),t)$. + +We need to show that this is a homotopy between $h\circ f$ and $k\circ f$. + +#### Theorem + +The Inclusion map $j:S^n\to \mathbb{R}^n-\{0\}$ induces on isomorphism of fundamental groups + +$$ +j_*:\pi_1(S^n)\to \pi_1(\mathbb{R}^n-\{0\}) +$$ + +The function is injective. + +> Recall we showed that $S^1\to \mathbb{R}-\{0\}$ is injective by $x\mapsto \frac{x}{|x|}$. + +We want to show that $j_*\circ r_*=id_{\pi_1(S^n)}\quad r_*\circ j_*=id_{\pi_1(\mathbb{R}^n-\{0\})}$, then $r_*$, $j_*$ are isomorphism. + +
+Proof + +**Homotopy is well defined**. + +Consider $H:(\mathbb{R}^n-\{0\})\times I\to \mathbb{R}^n-\{0\}$. + +Given $(x,t)\mapsto tx+(1-t)\frac{x}{\|x\|}$. + +Note that $(t-\frac{1-t}{\|x\|})x=0\implies t=0\land t=1$. + +So this map is well defined. + +**Base point is fixed**. + +On point $(1,0)$ (or anything on the sphere), $H(x,0)=x$. + +
+ +#### Definition of deformation retract + +Let $A$ be a subspace of $X$, we say that $A$ is a deformation retract of $X$ if the identity map of $X$ is homotopic to a map that carries all $X$ to $A$ such that each point of $A$ remains fixed during the homotopy. + +Equivalently, there exists a homotopy $H:X\times I\to X$ such that: + +- $H(x,0)=x$ forall $x\in X$ +- $H(a,t)=a$ for all $a\in A$, $t\in I$ +- $H(x,1)\in A$ for all $x\in X$ + +Equivalently, + +$r:H(x,1):X\to A$ is a retract. + +If we let $j:A\to X$ be the inclusion map, then $r\circ j=id_A$, and $j\circ r\sim id_X$ (with $A$ fixed.) + +
+Example of deformation retract + +$S^1$ is a deformation retract of $\mathbb{R}^2-\{0\}$ + +
+ +#### Theorem for Deformation Retract + +If $A$ is a deformation retract of $X$, then $A$ and $X$ have the same fundamental group. + diff --git a/content/Math4202/Math4202_L9.md b/content/Math4202/Math4202_L9.md index 1fee6fe..59d6064 100644 --- a/content/Math4202/Math4202_L9.md +++ b/content/Math4202/Math4202_L9.md @@ -7,7 +7,7 @@ Consider the space of paths up to homotopy equivalence. $$ -\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 $$ -(\Pi_1(X,x),*) +(pi_1(X,x),*) $$ \ No newline at end of file diff --git a/content/Math4202/_meta.js b/content/Math4202/_meta.js index a208511..a12baad 100644 --- a/content/Math4202/_meta.js +++ b/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)", } diff --git a/content/Math4302/Math4302_L25.md b/content/Math4302/Math4302_L25.md new file mode 100644 index 0000000..7b24419 --- /dev/null +++ b/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 + +
+Examples of division ring that is not a field + +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 $xx^{-1}=1$. + +
+ +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 $\operatorname{gcd}(m,n)=1$. + +
+Proof + +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 $[m]$. $mr\equiv 1\mod n\implies [m][x]=[1]$. + +
+ +### 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. + +
+Example of zero divisors + +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. + +$ab=0\implies a^{-1}ab=0\implies 1b=0\implies b=0$. + +
+ +> [!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). + +
+Proof + +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 $n=n_1d,m=m_1d, 1\leq n_1 + +#### Definition of integral domain + +A commutative ring with unity is called a integral domain (or just a domain) if it has no zero divisors. + +
+Example of integral domain + +$\mathbb{Z}$ is a integral domain. + +--- + +Any field is a integral domain. +
+ +#### 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. + +
+Proof + +if $a,b$ are units, then $ab$ is a unit $(ab)^{-1}=b^{-1}a^{-1}$. + +
+ +In particular, non-zero elements of any field form an abelian group under multiplication. + +
+Example + +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). +
diff --git a/content/Math4302/_meta.js b/content/Math4302/_meta.js index f5a4e1a..5406c99 100644 --- a/content/Math4302/_meta.js +++ b/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)", }