updates

content/Math4202/Math4202_L25.md

@@ -76,7 +76,7 @@ $$
 
 </details>
 
-#### Lemma
+#### Lemma of homotopy equivalence
 
 Let $f,g:X\to Y$ be continuous maps. let $f(x_0)=y_0$ and $g(x_0)=y_1$. If $f$ and $g$ are homotopic, then there is a path $\alpha:I\to Y$ such that $\alpha(0)=y_0$ and $\alpha(1)=y_1$.
 

content/Math4202/Math4202_L26.md

@@ -0,0 +1,90 @@
+# Math4202 Topology II (Lecture 26)
+
+## Algebraic Topology
+
+### Deformation Retracts and Homotopy Type
+
+#### Lemma of homotopy equivalence
+
+Let $f,g:X\to Y$ be continuous maps. let
+
+$$
+f_*=\pi_1(X,f(x_0))\quad\text{and}\quad g_*=\pi_1(Y,g(x_0))
+$$
+
+And $H:X\times I\to Y$ is a homotopy from $f$ to $g$ with a path $H(x_0,t)=\alpha(t)$ for all $t\in I$.
+
+Then $\hat{\alpha}\circ f_*=[\bar{\alpha}*(f\circ \gamma)*\alpha]=[g\circ \gamma]=g_*$. where $\gamma$ is a loop in $X$ based at $x_0$.
+
+<details>
+<summary>Proof</summary>
+$I\times I\xrightarrow{\gamma_{id}} X\times I\xrightarrow{H} Y$
+
+- $I\times \{0\}\mapsto f\circ\gamma$
+- $I\times \{1\}\mapsto g\circ\gamma$
+- $\{0\}\times I\mapsto \alpha$
+- $\{1\}\times I\mapsto \alpha$
+
+As $I\times I$ is convex, $I\times \{0\}\simeq (\{0\}\times I)*(I\times \{1\})*(\{1\}\times I)$.
+
+</details>
+
+#### Corollary for homotopic continuous maps
+
+Let $h,k$ be homotopic continuous maps. And let $h(x_0)=y_0,k(x_0)=y_1$. If $h_*:\pi_1(X,x_0)\to \pi_1(Y,y_0)$ is injective, then $k_*:\pi_1(X,x_0)\to \pi_1(Y,y_1)$ is injective.
+
+<details>
+<summary>Proof</summary>
+
+$\hat{\alpha}$ is an isomorphism of $\pi_1(Y,y_0)$ to $\pi_1(Y,y_1)$.
+
+</details>
+
+#### Corollary for nulhomotopic maps
+
+Let $h:X\to Y$ be nulhomotopic. Then $h_*:\pi_1(X,x_0)\to \pi_1(Y,h(x_0))$ is a trivial group homomorphism (mapping to the constant map on $h(x_0)$).
+
+#### Theorem for fundamental group isomorphism by homotopy equivalence
+
+Let $f:X\to Y$ be a continuous map. Let $f(x_0)=y_0$. If $f$ is a [homotopy equivalence](https://notenextra.trance-0.com/Math4202/Math4202_L25/#definition-of-homotopy-equivalence) ($\exists g:Y\to X$ such that $fg\simeq id_X$, $gf\simeq id_Y$), then
+
+$$
+f_*:\pi_1(X,x_0)\to \pi_1(Y,y_0)
+$$
+is an isomorphism.
+
+<details>
+<summary>Proof</summary>
+
+Let $g:Y\to X$ be the homotopy inverse of $f$. 
+
+Then,
+
+$f_*\circ g_*=\alpha \circ id_{\pi_1(Y,y_0)}=\alpha$
+
+And $g_*\circ f_*=\bar{\alpha}\circ id_{\pi_1(X,x_0)}=\bar{\alpha}$
+
+So $f_*\circ (g_*\circ \hat{\alpha}^-1)=id_{\pi_1(X,x_0)}$
+
+And $g_*\circ (f_*\circ \hat{\alpha}^-1)=id_{\pi_1(Y,y_0)}$
+
+So $f_*$ is an isomorphism (have left and right inverse).
+</details>
+
+### Fundamental group of higher dimensional sphere
+
+$\pi_1(S^n,x_0)=\{e\}$ for $n\geq 2$.
+
+We can decompose the sphere to the union of two hemisphere and compute $\pi_1(S^n_+,x_0)=\pi_1(S^n_-,x_0)=\{e\}$
+
+But for $n\geq 2$, $S^n_+\cap S^n_-=S^{n-1}$, where $S^1_+\cap S^1_-$ is two disjoint points.
+
+#### Theorem for "gluing" fundamental group
+
+Suppose $X=U\cup V$, where $U$ and $V$ are open subsets of $X$. Suppose that $U\cap V$ is path connected, and $x\in U\cap V$. Let $i,j$ be the inclusion maps of $U$ and $V$ into $X$, the images of the induced homomorphisms
+
+$$
+i_*:\pi_1(U,x_0)\to \pi_1(X,x_0)\quad j_*:\pi_1(V,x_0)\to \pi_1(X,x_0)
+$$
+
+The image of the two map generate $\pi_1(X,x_0)$.

content/Math4202/Math4202_L27.md

@@ -0,0 +1,69 @@
+# Math4202 Topology II (Lecture 27)
+
+## Algebraic Topology
+
+### Fundamental Groups for Higher Dimensional Sphere
+
+#### Theorem for "gluing" fundamental group
+
+Suppose $X=U\cup V$, where $U$ and $V$ are open subsets of $X$. Suppose that $U\cap V$ is path connected, and $x\in U\cap V$. Let $i,j$ be the inclusion maps of $U$ and $V$ into $X$, the images of the induced homomorphisms
+
+$$
+i_*:\pi_1(U,x_0)\to \pi_1(X,x_0)\quad j_*:\pi_1(V,x_0)\to \pi_1(X,x_0)
+$$
+
+The image of the two map generate $\pi_1(X,x_0)$.
+
+$G$ is a group, and let $S\subseteq G$, where $G$ is generated by $S$, if $\forall g\in G$, $\exists s_1,s_2,\ldots,s_n\in S$ such that $g=s_1s_2\ldots s_n\in G$. (We can write $G$ as a word of elements in $S$.)
+
+<details>
+<summary>Proof</summary>
+
+Let $f$ be a loop in $X$, $f\simeq g_1*g_2*\ldots*g_n$, where $g_i$ is a loop in $U$ or $V$.
+
+For example, consider the function, $f=f_1*f_2*f_3*f_4$, where $f_1\in S_+$, $f_2\in S_-$, $f_3\in S_+$, $f_4\in S_-$.
+
+Take the functions $\bar{\alpha_1}*\alpha_1\simeq e_{x_1}$ where $x_1$ is the intersecting point on $f_1$ and $f_2$.
+
+Therefore,
+
+$$
+\begin{aligned}
+f&=f_1*f_2*f_3*f_4\\
+&(f_1*\bar{\alpha})*(\alpha_1*f_2*\bar{\alpha_2})*(\alpha_2*f_3*\bar{\alpha_3})*(\alpha_4*f_4)
+\end{aligned}
+$$
+
+This decompose $f$ into a word of elements in either $S_+$ or $S_-$.
+
+---
+
+Note that $f$ is a continuous function $I\to X$, for $t\in I$, $\exists I_t$ being a small neighborhood of $t$ such that $f(I_t)\subseteq U$ or $f(I_t)\subseteq V$.
+
+Since $U_{t\in I}I_t=I$, then $\{I_t\}_{t\in I}$ is an open cover of $I$.
+
+By compactness of $I$, there is a finite subcover $\{I_{t_1},\ldots,I_{t_n}\}$.
+
+Therefore, we can create a partition of $I$ into $[s_i,s_{i+1}]\subseteq I_{t_k}$ for some $k$.
+
+Then with the definition of $I_{t_k}$, $f([s_i,s_{i+1}])\subseteq U$ or $V$.
+
+Then we can connect $x_0$ to $f(s_i)$ with a path $\alpha_i\subseteq U\cap V$.
+
+$$
+\begin{aligned}
+f&=f|_{[s_0,s_1]}*f|_{[s_1,s_2]}*\ldots**f|_{[s_{n-1},s_n]}\\
+&\simeq f|_{[s_0,s_1]}*(\bar{\alpha_1}*\alpha_1)*f|_{[s_1,s_2]}*(\bar{\alpha_2}*\alpha_2)*\ldots*f|_{[s_{n-1},s_n]}*(\bar{\alpha_n}*\alpha_n
+)\\
+&=(f|_{[s_0,s_1]}*\bar{\alpha_1})*(\alpha_1*f|_{[s_1,s_2]}*\bar{\alpha_2})*\ldots*(\alpha_{n-1}*f|_{[s_{n-1},s_n]}*\bar{\alpha_n})\\
+&=g_1*g_2*\ldots*g_n
+\end{aligned}
+$$
+
+</details>
+
+#### Corollary in higher dimensional sphere
+
+Since $S^n_+$ and $S^n_-$ are homeomorphic to open balls $B^n$, then $\pi_1(S^n_+,x_0)=\pi_1(S^n_-,x_0)=\pi_1(B^n,x_0)=\{e\}$ for $n\geq 2$.
+
+> Preview: Van Kampen Theorem

content/Math4202/_meta.js

@@ -31,4 +31,6 @@ export default {
     Math4202_L23: "Topology II (Lecture 23)",
     Math4202_L24: "Topology II (Lecture 24)",
     Math4202_L25: "Topology II (Lecture 25)",
+    Math4202_L26: "Topology II (Lecture 26)",
+    Math4202_L27: "Topology II (Lecture 27)",
 }

content/Math4302/Math4302_L26.md

@@ -2,7 +2,7 @@
 
 ## Rings
 
-### Integral Domains
+### Fermat’s and Euler’s Theorems
 
 Recall from last lecture, we consider $\mathbb{Z}_p$ and $\mathbb{Z}_p^*$ denote the group of units in $\mathbb{Z}_p$ with multiplication.
 
@@ -104,7 +104,7 @@ Solution for $2x\equiv 1\mod 3$
 
 So solution for $2x\equiv 1\mod 3$ is $\{3k+2|k\in \mathbb{Z}\}$.
 
-#### Theorem for solving modular equations
+#### Theorem for exsistence of solution of modular equations
 
 $ax\equiv b\mod n$ has a solution if and only if $\operatorname{gcd}(a,n)|b$ and in that case the equation has $d$ solutions in $\mathbb{Z}_n$.
 

content/Math4302/Math4302_L27.md

@@ -0,0 +1,126 @@
+# Math4302 Modern Algebra (Lecture 27)
+
+## Rings
+
+### Fermat’s and Euler’s Theorems
+
+Recall from last lecture, $ax\equiv b \mod n$, if $x\equiv y\mod n$, then $x$ is a solution if and only if $y$ is a solution.
+
+#### Theorem for existence of solution of modular equations
+
+$ax\equiv b\mod n$ has a solution if and only if $d=\operatorname{gcd}(a,n)|b$ And if there is a solution, then there are exactly $d$ solutions in $\mathbb{Z}_n$.
+
+<details>
+<summary>Proof</summary>
+
+For the forward direction, we proved if $ax\equiv b\mod n$ then $ax-b=ny$, $y\in\mathbb{Z}$.
+
+then $b=ax-ny$, $d|(ax-ny)$ implies that $d|b$.
+
+---
+
+For the backward direction, assume $d=\operatorname{gcd}(a,n)=1$. Then we need to show, there is exactly $1$ solution between $0$ and $n-1$.
+
+If $ax\equiv b\mod n$, then in $\mathbb{Z}_n$, $[a][x]=[b]$. (where $[a]$ denotes the remainder of $a$ by $n$ and $[b]$ denotes the remainder of $b$ by $n$)
+
+Since $\operatorname{gcd}(a,n)=1$, then $[a]$ is a unit in $\mathbb{Z}_n$, so we can multiply the above equation by the inverse of $[a]$. and get $[x]=[a]^{-1}[b]$.
+
+Now assume $d=\operatorname{gcd}(a,n)$ where $n$ is arbitrary. Then $a=a'd$, then $n=n'd$, with $\operatorname{gcd}(a',n')=1$.
+
+Also $d|b$ so $b=b'd$. So
+
+$$
+\begin{aligned}
+ax\equiv b \mod n&\iff n|(ax-b)\\
+&\iff n'd|(a'dx-b'd)\\
+&\iff n'|(a'x-b')\\
+&\iff a'x\equiv b'\mod n'
+\end{aligned}
+$$.
+
+Since $\operatorname{gcd}(a',n')=1$, there is a unique solution $x_0\in \mathbb{Z}_{n'}$. $0\leq x_0\leq n'+1$. Other solution in $\mathbb{Z}$ are of the form $x_0+kn'$ for $k\in \mathbb{Z}$.
+
+And there will be $d$ solutions in $\mathbb{Z}_n$,
+
+</details>
+
+<details>
+<summary>Examples</summary>
+
+Solve $12x\equiv 25\mod 7$.
+
+$12\equiv 5\mod 7$, $25\equiv 4\mod 7$. So the equation becomes $5x\equiv 4\mod 7$.
+
+$[5]^{-1}=3\in \mathbb{Z}_7$, so $[5][x]\equiv [4]$ implies $[x]\equiv [3][4]\equiv [5]\mod 7$.
+
+So solution in $\mathbb{Z}$ is $\{5+7k:k\in \mathbb{Z}\}$.
+
+---
+
+Solve $6x\equiv 32\mod 20$.
+
+$\operatorname{gcd}(6,20)=2$, so $6x\equiv 12\mod 20$ if and only if $3x\equiv 6\mod 10$.
+
+$[3]^{-1}=[7]\in \mathbb{Z}_{10}$, so $[3][x]\equiv [6]$ implies $[x]\equiv [7][6]\equiv [2]\mod 10$.
+
+So solution in $\mathbb{Z}_{20}$ is $[2]$ and $[12]$
+
+So solution in $\mathbb{Z}$ is $\{2+10k:k\in \mathbb{Z}\}$
+
+</details>
+
+### Ring homomorphisms
+
+#### Definition of ring homomorphism
+
+Let $R,S$ be two rings, $f:R\to S$ is a ring homomorphism if $\forall a,b\in R$,
+
+- $f(a+b)=f(a)+f(b)\implies f(0)=0, f(-a)=-f(a)$
+- $f(ab)=f(a)f(b)$
+
+#### Definition of ring isomorphism
+
+If $f$ is a ring homomorphism and a bijection, then $f$ is called a ring isomorphism.
+
+<details>
+<summary>Example</summary>
+Let $f:(\mathbb{Z},+,\times)\to(2\mathbb{Z},+,\times)$ by $f(a)=2a$.
+
+Is not a ring homomorphism since $f(ab)\neq f(a)f(b)$ in general.
+
+---
+
+Let $f:(\mathbb{Z},+,\times)\to(\mathbb{Z}_n,+,\times)$ by $f(a)=a\mod n$
+
+Is a ring homomorphism.
+
+</details>
+
+### Integral domains and their file fo fractions.
+
+Let $R$ be an integral domain: (i.e. $R$ is commutative with unity and no zero divisors).
+
+#### Definition of field of fractions
+
+If $R$ is an integral domain, we can construct a field containing $R$ called the field of fractions (or called field of quotients) of $R$.
+
+$$
+S=\{(a,b)|a,b\in R, b\neq 0\}
+$$
+
+a relation on $S$ is defined as follows:
+
+$(a,b)\sim (c,d)$ if and only if $ad=bc$.
+
+<details>
+<summary>This equivalence relation is well defined</summary>
+
+- Reflectivity: $(a,b)\sim (a,b)$ $ab=ab$
+- Symmetry: $(a,b)\sim (c,d)\Rightarrow (c,d)\sim (a,b)$
+- Transitivity: $(a,b)\sim (c,d)$ and $(c,d)\sim (e,f)\Rightarrow (a,b)\sim (e,f)$
+  - $ad=bc$, and $cf=ed$, we want to conclude that $af=be$. since $ad=bc$, then $adf=bcf$, since $cf=ed$, then $cfb=edb$, therefore $adf=edb$.
+  - Then $d(af-be)=0$ since $d\neq 0$ then $af=be$.
+
+</details>
+
+Then $S/\sim$ is a field.

content/Math4302/Math4302_L28.md

@@ -0,0 +1,153 @@
+# Math4302 Modern Algebra (Lecture 28)
+
+## Rings
+
+### Field of quotients
+
+Let $R$ be an integral domain ($R$ has unity and commutative with no zero divisors).
+
+Consider the pair $S=\{(a,b)|a,b\in R, b\neq 0\}$.
+
+And define the equivalence relation on $S$ as follows:
+
+$(a,b)\sim (c,d)$ if and only if $ad=bc$.
+
+We denote $[(a,b)]$ as set of all elements in $S$ equivalent to $(a,b)$.
+
+Let $F$ be the set of all equivalent classes. We define addition and multiplication on $F$ as follows:
+
+$$
+[(a,b)]+[(c,d)]=[(ad+bc,bd)]
+$$
+
+$$
+[(a,b)]\cdot[(c,d)]=[(ac,bd)]
+$$
+
+<details>
+<summary>The multiplication and addition is well defined </summary>
+
+Addition:
+
+If $(a,b)\sim (a',b')$, and $(c,d)\sim (c',d')$, then we want to show that $(ad+bc,bd)\sim (a'd+c'd,b'd)$.
+
+Since $(a,b)\sim (a',b')$, then $ab'=a'b$; $(c,d)\sim (c',d')$, then $cd'=dc'$,
+
+So $ab'dd'=a'bdd'$, and $cd'bb'=dc'bb'$.
+
+ $adb'd'+bcb'd'=a'd'bd+b'c'bd$, therefore $(ad+bc,bd)\sim (a'd+c'd,b'd)$.
+
+---
+
+Multiplication:
+
+If $(a,b)\sim (a',b')$, and $(c,d)\sim (c',d')$, then we want to show that $(ac,bd)\sim (a'c',b'd')$.
+
+Since $(a,b)\sim (a',b')$, then $ab'=a'b$; $(c,d)\sim (c',d')$, then $cd'=dc'$, so $(ac,bd)\sim (a'c',b'd')$
+
+</details>
+
+#### Claim (F,+,*) is a field
+
+- additive identity: $(0,1)\in F$
+- additive inverse: $(a,b)\in F$, then $(-a,b)\in F$ and $(-a,b)+(a,b)=(0,1)\in F$
+- additive associativity: bit long.
+
+- multiplicative identity: $(1,1)\in F$
+- multiplicative inverse: $[(a,b)]$ is non zero if and only if $a\neq 0$, then $a^{-1}=[(b,a)]\in F$.
+- multiplicative associativity: bit long
+
+- distributivity: skip, too long.
+
+Such field is called a quotient field of $R$.
+
+And $F$ contains $R$ by $\phi:R\to F$, $\phi(a)=[(a,1)]$.
+
+This is a ring homomorphism.
+
+- $\phi(a+b)=[(a+b,1)]=[(a,1)][(b,1)]\phi(a)+\phi(b)$
+- $\phi(ab)=[(ab,1)]=[(a,1)][(b,1)]\phi(a)\phi(b)$
+
+and $\phi$ is injective.
+
+If $\phi(a)=\phi(b)$, then $a=b$.
+
+<details>
+<summary>Example</summary>
+
+Let $D\subset \mathbb R$ and
+
+$$
+\mathbb Z \subset D\coloneqq \{a+b\sqrt{2}:a,b\in \mathbb Z\}
+$$
+
+Then $D$ is a subring of $\mathbb R$, and integral domain, with usual addition and multiplication.
+
+$$
+(a+b\sqrt{2})(c+d\sqrt{2})=(ac+2bd)+(ad+bc)\sqrt{2}
+$$
+
+$$
+-(a+b\sqrt{2})=(-a)+(-b)\sqrt{2})
+$$
+
+...
+
+$D$ is a integral domain since $\mathbb R$ has no zero divisors, therefore $D$ has no zero divisors.
+
+Consider the field of quotients of $D$. $[(a+b\sqrt{2},c+d\sqrt{2})]$. This is isomorphic to $\mathbb Q(\sqrt2)=\{r+s\sqrt{2}:r,s\in \mathbb Q\}$
+
+$$
+m+n\sqrt{2}=\frac{m}{n}+\frac{m'}{n'}\sqrt{2}\mapsto [(mn'+nm'\sqrt{2},nn')]
+$$
+
+And use rationalization on the forward direction.
+
+</details>
+
+#### Polynomial rings
+
+Let $R$ be a ring, a polynomial with coefficients in $R$ is a sum
+
+$$
+a_0+a_1x+\cdots+a_nx^n
+$$
+
+where $a_i\in R$. $x$ is indeterminate, $a_0,a_1,\cdots,a_n$ are called coefficients. $a_0$ is the constant term.
+
+If $f$ is a non-zero polynomial, then the degree of $f$ is defined as the largest $n$ such that $a_n\neq 0$.
+
+<details>
+<summary>Example</summary>
+
+Let $f=1+2x+0x^2-1x^3+0x^4$, then $deg f=3$
+
+</details>
+
+If $R$ has a unity $1$, then we write $x^m$ instead of $1x^m$.
+
+Let $R[x]$ denote the set of all polynomials with coefficients in $R$.
+
+We define multiplication and addition on $R[x]$.
+
+$f:a_0+a_1x+\cdots+a_nx^n$
+
+$g:b_0+b_1x+\cdots+b_mx^m$
+
+Define,
+
+$$
+f+g=a_0+b_0+a_1x+b_1x+\cdots+a_nx^n+b_mx^m
+$$
+
+$$
+fg=(a_0b_0)+(a_1b_0)x+\cdots+(a_nb_m)x^m
+$$
+
+In general, the coefficient of $x^m=\sum_{i=0}^{m}a_ix^{m-i}$.
+
+> [!CAUTION]
+>
+> The field $R$ may not be commutative, follow the order of computation matters.
+
+We will show that this is a ring and explore additional properties.

content/Math4302/_meta.js

@@ -29,4 +29,6 @@ export default {
     Math4302_L24: "Modern Algebra (Lecture 24)",
     Math4302_L25: "Modern Algebra (Lecture 25)",
     Math4302_L26: "Modern Algebra (Lecture 26)",
+    Math4302_L27: "Modern Algebra (Lecture 27)",
+    Math4302_L28: "Modern Algebra (Lecture 28)",
 }