Merge branch 'main' of https://git.trance-0.com/Trance-0/NoteNextra-origin

content/Math4202/Math4202_L23.md

@@ -94,3 +94,48 @@ For this section, we will show that $h_*$ is an isomorphism.
 #### Lemma for equality of homomorphism
 
 Let $h,k: (X,x_0)\to (Y,y_0)$ be continuous maps. If $h$ and $k$ are homotopic, and if **the image of $x_0$ under the homotopy remains $y_0$**. The homomorphism $h_*$ and $k_*$ from $\pi_1(X,x_0)$ to $\pi_1(Y,y_0)$ are equal.
+
+<details>
+<summary>Proof</summary>
+
+Let $H:X\times I\to Y$ be a homotopy from $h$ to $k$ such that
+$$
+H(x,0)=h(x), \qquad H(x,1)=k(x), \qquad H(x_0,t)=y_0 \text{ for all } t\in I.
+$$
+
+To show $h_*=k_*$, let $[f]\in \pi_1(X,x_0)$ be arbitrary, where
+$f:I\to X$ is a loop based at $x_0$, so $f(0)=f(1)=x_0$.
+
+Define
+$$
+F:I\times I\to Y,\qquad F(s,t)=H(f(s),t).
+$$
+Since $H$ and $f$ are continuous, $F$ is continuous. For each fixed $t\in I$, the map
+$$
+s\mapsto F(s,t)=H(f(s),t)
+$$
+is a loop based at $y_0$, because
+$$
+F(0,t)=H(f(0),t)=H(x_0,t)=y_0
+\quad\text{and}\quad
+F(1,t)=H(f(1),t)=H(x_0,t)=y_0.
+$$
+Thus $F$ is a based homotopy between the loops $h\circ f$ and $k\circ f$, since
+$$
+F(s,0)=H(f(s),0)=h(f(s))=(h\circ f)(s),
+$$
+and
+$$
+F(s,1)=H(f(s),1)=k(f(s))=(k\circ f)(s).
+$$
+
+Therefore $h\circ f$ and $k\circ f$ represent the same element of $\pi_1(Y,y_0)$, so
+$$
+[h\circ f]=[k\circ f].
+$$
+Hence
+$$
+h_*([f])=[h\circ f]=[k\circ f]=k_*([f]).
+$$
+Since $[f]$ was arbitrary, it follows that $h_*=k_*$.
+</details>

content/Math4202/Math4202_L24.md

@@ -70,6 +70,12 @@ If we let $j:A\to X$ be the inclusion map, then $r\circ j=id_A$, and $j\circ r\s
 
 $S^1$ is a deformation retract of $\mathbb{R}^2-\{0\}$ 
 
+---
+
+Consider $\mathbb{R}^2-p=q$, the doubly punctured plane. "The figure 8" space is the deformation retract.
+
+![Retraction of doubly punctured plane](https://notenextra.trance-0.com/Math4202/Retraction_of_doubly_punctured_plane.jpg)
+
 </details>
 
 #### Theorem for Deformation Retract

content/Math4202/Math4202_L25.md

@@ -0,0 +1,89 @@
+# Math4202 Topology II (Lecture 25)
+
+## Algebraic Topology
+
+### Deformation Retracts and Homotopy Type
+
+Recall from last lecture, Let $A\subseteq X$, if there exists a continuous map (deformation retraction) $H:X\times I\to X$ such that
+
+- $H(x,0)=x$ for all $x\in X$
+- $H(x,1)\in A$ for all $x\in X$
+- $H(a,t)=a$ for all $a\in A$, $t\in I$
+
+then the inclusion map$\pi_1(A,a)\to \pi_1(X,a)$ is an isomorphism.
+
+<details>
+<summary>Example for more deformation retract</summary>
+
+Let $X=\mathbb{R}^3-\{0,(0,0,1)\}$.
+
+Then the two sphere with one point intersect is a deformation retract of $X$.
+
+---
+
+Let $X$ be $\mathbb{R}^3-\{(t,0,0)\mid t\in \mathbb{R}\}$, then the cyclinder is a deformation retract of $X$.
+
+</details>
+
+#### Definition of homotopy equivalence
+
+Let $f:X\to Y$ and $g:Y\to X$ be a continuous maps.
+
+Suppose 
+
+- the map $g\circ f:X\to X$ is homotopic to the identity map $\operatorname{id}_X$.
+- the map $f\circ g:Y\to Y$ is homotopic to the identity map $\operatorname{id}_Y$.
+
+Then $f$ and $g$ are **homotopy equivalences**, and each is said to be the **homotopy inverse** of the other.
+
+$X$ and $Y$ are said to be **homotopy equivalent**.
+
+<details>
+<summary>Example</summary>
+
+Consider the punctured torus $X=S^1\times S^1-\{(0,0)\}$.
+
+Then we can do deformation retract of the glued square space to boundary of the square.
+
+After glueing, we left with the figure 8 space.
+
+Then $X$ is homotopy equivalent to the figure 8 space.
+
+</details>
+
+Recall the lemma, [Lemma for equality of homomorphism](https://notenextra.trance-0.com/Math4202/Math4202_L23/#lemma-for-equality-of-homomorphism)
+
+Let $f:X\to Y$ and $g:X\to Y$, with homotopy $H:X\times I\to Y$, such that
+
+- $H(x,0)=f(x)$ for all $x\in X$
+- $H(x,1)=g(x)$ for all $x\in X$
+- $H(x,t)=y_0$ for all $t\in I$, and $y_0\in Y$ is fixed.
+
+Then $f_*=g_*:\pi_1(X,x_0)\to \pi_1(Y,y_0)$ is an isomorphism.
+
+We wan to know if it is safe to remove the assumption that $y_0$ is fixed.
+
+<details>
+<summary>Idea of Proof</summary>
+
+Let $k$ be any loop in $\pi_1(X,x_0)$.
+
+We can correlate the two fundamental group $f\cric k$ by the function $\alpha:I\to Y$, and $\hat{\alpha}:\pi_1(Y,y_0)\to \pi_1(Y,y_1)$. (suppose $f(x_0)=y_0, g(x_0)=y_1$), it is sufficient to show that
+
+$$
+f\circ k\simeq \alpha *(g\circ k)*\bar{\alpha}
+$$
+
+</details>
+
+#### Lemma
+
+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 $\alpha(1)=y_1$.
+
+Defined as the restriction of the homotopy to $\{x_0\}\times I$, satisfying $\hat{\alpha}\circ f_*=g_*$.
+
+Imagine a triangle here:
+
+- $\pi_1(X,x_0)\to \pi_1(Y,y_0)$ by $f_*$
+- $\pi_1(Y,y_0)\to \pi_1(Y,y_1)$ by $\hat{\alpha}$
+- $\pi_1(Y,y_1)\to \pi_1(X,x_0)$ by $g_*$

content/Math4202/_meta.js

@@ -30,4 +30,5 @@ export default {
     Math4202_L22: "Topology II (Lecture 22)",
     Math4202_L23: "Topology II (Lecture 23)",
     Math4202_L24: "Topology II (Lecture 24)",
+    Math4202_L25: "Topology II (Lecture 25)",
 }

content/Math4302/Math4302_L26.md

@@ -0,0 +1,111 @@
+# Math4302 Modern Algebra (Lecture 26)
+
+## Rings
+
+### Integral Domains
+
+Recall from last lecture, we consider $\mathbb{Z}_p$ and $\mathbb{Z}_p^*$ denote the group of units in $\mathbb{Z}_p$ with multiplication.
+
+$$
+\mathbb{Z}_p^* = \{1,2,\cdots,p-1\}, \quad |\mathbb{Z}_p^*| = p-1
+$$
+
+Let $[a]\in \mathbb{Z}_p^*$, then $[a]^{p-1}=[1]$, this implies that $a^{p-1}\mod p=1$. 
+
+Now if $m\in \mathbb{Z}$ and $a=$ remainder of $m$ by $p$, $[a]\in \mathbb{Z}_p$, implies $m\equiv a\mod p$.
+
+Then $m^{p-1}\equiv a^{p-1}\mod p$.
+
+So
+
+#### Fermat’s little theorem
+
+If $p$ is not a divisor of $m$, then $m^{p-1}\equiv 1\mod p$.
+
+#### Corollary of Fermat’s little theorem
+
+If $m\in \mathbb{Z}$, then $m^p\equiv m\mod p$.
+
+<details>
+<summary>Proof</summary>
+
+If $p|m$, then $m^{p-1}\equiv 0\equiv m\mod p$.
+
+If $p\not|m$, then by Fermat’s little theorem, $m^{p-1}\equiv 1\equiv m\mod p$, so $m^p\equiv m\mod p$.
+
+</details>
+
+<details>
+<summary>Example</summary>
+
+Find the remainder of $40^{100}$ by $19$.
+
+$40^{100}\equiv 2^{100}\mod 19$
+
+$2^{100}\equiv 2^{10}\mod 19$ (Fermat’s little theorem $2^18\equiv 1\mod 19, 2^{90}\equiv 1\mod 19$)
+
+$2^10\equiv (-6)^2\mod 19\equiv 36\mod 19\equiv 17\mod 19$
+
+---
+
+For every integer $n$, $15|(n^{33}-n)$.
+
+$15=3\cdot 5$, therefore enough to show that $3|(n^{33}-n)$ and $5|(n^{33}-n)$.
+
+Apply the corollary of Fermat’s little theorem to $p=3$: $n^3\equiv n\mod 3$, $(n^3)^11\equiv n^{11}\equiv (n^3)^3 n^2=n^3 n^2\equiv n^3\mod 3\equiv n\mod 3$.
+
+Therefore $3|(n^{33}-n)$.
+
+Apply the corollary of Fermat’s little theorem to $p=5$: $n^5\equiv n\mod 5$, $n^30 n^3\equiv (n^5)^6 n^3\equiv n^6 n^3\equiv n^5\mod 5\equiv n\mod 5$.
+
+Therefore $5|(n^{33}-n)$.
+
+</details>
+
+#### Euler’s totient function
+
+Consider $\mathbb{Z}_6$, by definition for the group of units, $\mathbb{Z}_6^*=\{1,5\}$.
+
+$$
+\phi(n)=|\mathbb{Z}_n^*|=|\{1\leq x\leq n:gcd(x,n)=1\}|
+$$
+
+<details>
+<summary>Example</summary>
+
+$\phi(8)=|\{1,3,5,7\}|=4$
+
+</details>
+
+If $[a]\in \mathbb{Z}_n^*$, then $[a]^{\phi(n)}=[1]$. So $a^{\phi(n)}\equiv 1\mod n$.
+
+#### Theorem
+
+If $m\in \mathbb{Z}$, and $gcd(m,n)=1$, then $m^{\phi(n)}\equiv 1\mod n$.
+
+<details>
+<summary>Proof</summary>
+
+If $a$ is the remainder of $m$ by $n$, then $m\equiv a\mod n$, and $\operatorname{gcd}(a,n)=1$, so $m^{\phi(n)}\equiv a^{\phi(n)}\equiv 1\mod n$.
+
+</details>
+
+#### Applications on solving modular equations
+
+Solving equations of the form $ax\equiv b\mod n$.
+
+Not always have solution, $2x\equiv 1\mod 4$ has no solution since $1$ is odd.
+
+Solution for $2x\equiv 1\mod 3$
+
+- $x\equiv 0\implies 2x\equiv 0\mod 3$
+- $x\equiv 1\implies 2x\equiv 2\mod 3$
+- $x\equiv 2\implies 2x\equiv 1\mod 3$
+
+So solution for $2x\equiv 1\mod 3$ is $\{3k+2|k\in \mathbb{Z}\}$.
+
+#### Theorem for solving modular equations
+
+$ax\equiv b\mod n$ has a solution if and only if $\operatorname{gcd}(a,n)|b$ and in that case the equation has $d$ solutions in $\mathbb{Z}_n$.
+
+Proof on next lecture.

content/Math4302/_meta.js

@@ -28,4 +28,5 @@ export default {
     Math4302_L23: "Modern Algebra (Lecture 23)",
     Math4302_L24: "Modern Algebra (Lecture 24)",
     Math4302_L25: "Modern Algebra (Lecture 25)",
+    Math4302_L26: "Modern Algebra (Lecture 26)",
 }