diff --git a/content/Math4202/Math4202_L23.md b/content/Math4202/Math4202_L23.md index 8a953b5..4ff60f8 100644 --- a/content/Math4202/Math4202_L23.md +++ b/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. + +
+Proof + +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 $h_*=k_*$. +
diff --git a/content/Math4202/Math4202_L24.md b/content/Math4202/Math4202_L24.md index 1280e1c..cd89558 100644 --- a/content/Math4202/Math4202_L24.md +++ b/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) + #### Theorem for Deformation Retract diff --git a/content/Math4202/Math4202_L25.md b/content/Math4202/Math4202_L25.md new file mode 100644 index 0000000..11492f1 --- /dev/null +++ b/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. + +
+Example for more deformation retract + +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 $X$. + +
+ +#### 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**. + +
+Example + +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. + +
+ +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. + +
+Idea of Proof + +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 + +$$ +f\circ k\simeq \alpha *(g\circ k)*\bar{\alpha} +$$ + +
+ +#### 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 $g_*$ diff --git a/content/Math4202/_meta.js b/content/Math4202/_meta.js index a12baad..63ad3d7 100644 --- a/content/Math4202/_meta.js +++ b/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)", } diff --git a/content/Math4302/Math4302_L26.md b/content/Math4302/Math4302_L26.md new file mode 100644 index 0000000..186b33d --- /dev/null +++ b/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 $m^p\equiv m\mod p$. + +
+Proof + +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 $m^p\equiv m\mod p$. + +
+ +
+Example + +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 $5|(n^{33}-n)$. + +
+ +#### Euler’s totient function + +Consider $\mathbb{Z}_6$, by definition for the group of units, $\mathbb{Z}_6^*=\{1,5\}$. + +$$ +\phi(n)=|\mathbb{Z}_n^*|=|\{1\leq x\leq n:gcd(x,n)=1\}| +$$ + +
+Example + +$\phi(8)=|\{1,3,5,7\}|=4$ + +
+ +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 $m^{\phi(n)}\equiv 1\mod n$. + +
+Proof + +If $a$ is the remainder of $m$ by $n$, then $m\equiv a\mod n$, and $\operatorname{gcd}(a,n)=1$, so $m^{\phi(n)}\equiv a^{\phi(n)}\equiv 1\mod n$. + +
+ +#### 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. \ No newline at end of file diff --git a/content/Math4302/_meta.js b/content/Math4302/_meta.js index 5406c99..830e780 100644 --- a/content/Math4302/_meta.js +++ b/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)", } diff --git a/public/Math4202/Retraction_of_doubly_punctured_plane.jpg b/public/Math4202/Retraction_of_doubly_punctured_plane.jpg new file mode 100644 index 0000000..8b1315e Binary files /dev/null and b/public/Math4202/Retraction_of_doubly_punctured_plane.jpg differ