updates

content/Math4202/Math4202_L20.md

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+# Math4202 Topology II (Lecture 20)
+
+## Algebraic Topology
+
+### Retraction and fixed point
+
+#### Lemma of retraction
+
+Let $h:S^1\to X$ be a continuous map. The following are equivalent:
+
+- $h$ is null-homotopic ($h$ is homotopic to a constant map).
+- $h$ extends to a continuous map $k:B_1(0)\to X$. ($k|_{\partial B_1(0)}=h$)
+
+> For this course, we use closed disk $B_1(0)=\{(x,y)|d((x,y),(0,0))\leq 1\}$.
+
+- $h_*$ is the trivial group homomorphism of fundamental groups (Image of $\pi_1(S^1,x_0)\to \pi_1(X,x_0)$ is trivial group, identity).
+
+<details>
+<summary>Proof</summary>
+
+First we will show that (1) implies (2).
+
+By the null homotopic condition, there exists $H:S\times I\to X$ that $H(s,1)=h(s),H(s,0)=x_0$.
+
+Define the equivalence relation for all the point in the set $H(s,0)$. Then there exists $\tilde{H}:S\times I/\sim \to X$ is a continuous map. (by quotient map $q:S\times I\to S\times I/\sim$ and $H$ is continuous.)
+
+Note that the cone shape is homotopic equivalent to the disk using polar coordinates. $k|_{\partial B_1(0)}=H|_{S^1\times\{1\}}=h$.
+
+---
+
+Then we will prove that (2) implies (3).
+
+Let $i: S^1\to B_1(0)$ be the inclusion map. Then $k\circ i=h$, $k_*\circ i_*=h_*:\pi_1(S^1,1)\to \pi_1(X,h(1))$.
+
+Recall that $k:B_1(0)\to X$, $k_*:\pi_1(B_1(0),0)\to \pi_1(X,k(0))$ is trivial, since $B_1(0)$ is contractible.
+
+Therefore $k_*\circ i_*=h_*$ is the trivial group homomorphism.
+
+---
+
+Now we will show that (3) implies (1).
+
+Consider the map from $\alpha:[0,1]\to S^1$ by $\alpha:x\mapsto e^{2\pi ix}$. $[\alpha]$ is a generator of $\pi_1(S^1,1)$. (The lifting of $\alpha$ to $\mathbb{R}$ is $1$, which is a generator of $\mathbb{Z}$.)
+
+$h\circ \alpha:[0,1]\to X$ is a loop in $X$ representing $h_*([\alpha])$.
+
+As $h_*([\alpha])$ is trivial, $h\circ \alpha$ is homotopic to a constant loop.
+
+Therefore, there exist a homotopy $H:I\times I\to X$, where $H(s,0)=$ constant map, $H(s,1)=h\circ \alpha(s)$.
+
+Take $\tilde{H}:S\times I/\sim \to X$ by $\tilde{H}(\exp(2\pi is),0)=H(s,0)$, $\tilde{H}(\exp(2\pi is),t)=H(s,t)$. $x\in [0,1]$.
+
+(From the perspective of quotient map, we can see that $\alpha$ is the quotient map from $I\times I$ to $I\times I/\sim=S\times I$. Then $\tilde{H}$ is the induced continuous map from $S\times I$ to $X$.)
+
+</details>
+
+#### Corollary of punctured plane
+
+$i:S^1\to \mathbb{R}^2-\{0\}$ is not null homotopic.
+
+<details>
+<summary>Proof</summary>
+
+Recall from last lecture, $r:\mathbb{R}^2-\{0\}\to S^1$ is the retraction $x\mapsto \frac{x}{|x|}$.
+
+Therefore, we have $i_*:\pi_1(S^1,1)\to \pi_1(\mathbb{R}^2-\{0\},r(0))$: $\mathbb{Z}\to \pi_1(\mathbb{R}^2-\{0\},r(0))$ is injective.
+
+Therefore $i_*$ is non trivial.
+
+Therefore $i$ is not null homotopic.
+
+</details>

content/Math4202/_meta.js

@@ -22,4 +22,5 @@ export default {
     Math4202_L17: "Topology II (Lecture 17)",
     Math4202_L18: "Topology II (Lecture 18)",
     Math4202_L19: "Topology II (Lecture 19)",
+    Math4202_L20: "Topology II (Lecture 20)",
 }

content/Math4302/Math4302_L19.md

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+# Math4302 Modern Algebra (Lecture 19)
+
+## Solution for Midterm I
+
+Not applicable
+
+## Group
+
+Side notes: From previous lecture
+
+What is the group $\mathbb{Z}\times \mathbb{Z}/\langle (a,b)\rangle$, where $a,b\in \mathbb{Z}^+$?
+
+This should be isomorphic to $\mathbb{Z}\times \mathbb{Z}_{\operatorname{gcd}(a,b)}$
+
+<details>
+<summary>Proof</summary>
+
+If $\operatorname{gcd}(a,b)=1$, then $\mathbb{Z}\times \mathbb{Z}/\langle (a,b)\rangle\simeq \mathbb{Z}$.
+
+The general isomorphism $\phi:\mathbb{Z}\times \mathbb{Z}/\langle (a,b)\rangle\to \mathbb{Z}\times \mathbb{Z}_{\operatorname{gcd}(a,b)}$ is given by
+
+$$
+(a,b)\mapsto (\frac{ma-nb}{d},au+bv\mod d)
+$$
+
+where $u,v$ are two integers such that $mu+nv=d$.
+
+Geometrically, partition the lattice to grids then map each element in grid to continuous integer in $ab/\operatorname{gcd}(a,b)=\operatorname{lcm}(a,b)$.
+
+$\phi$ is surjective homomorphism, and $\ker(\phi)=\langle (a,b)\rangle$
+
+</details>
+
+### Center of a group
+
+Recall from previous lecture, the center of a group $G$ is the subgroup of $G$ that contains all elements that commute with all elements in $G$.
+
+$$
+Z(G)=\{a\in G\mid \forall g\in G, ag=ga\}
+$$
+
+this subgroup is normal and measure the "abelian" for a group.
+
+<details>
+<summary>Example</summary>
+
+$Z(S_3)=\{e\}$, all the transpositions are not commutative, so $Z(S_3)=\{e\}$.
+
+---
+
+$Z(GL_n(\mathbb{R}))=\{A\in GL_n(\mathbb{R})\mid AB=BA\text{ for all }B\in GL_n(\mathbb{R})\}$
+
+This is all the multiples of the identity matrix.
+
+Ideas of proof:
+
+1. Consider all matrix that is have non-zero element other than diagonal, we can always find a matrix that don't commute with it.
+2. Consider if the diagonal entry are different, the row flip matrix will make your life worse.
+
+Therefore the only choice left the multiple of the identity matrix.
+
+</details>
+
+#### Definition of the commutator of a group
+
+Let $G$ be a group and $a,b\in G$, the commutator $[a,b]$ is defined as $aba^{-1}b^{-1}$.
+
+$[a,b]=e$ if and only if $a$ and $b$ commute.
+
+Some additional properties:
+
+- $[a,b]^{-1}=[b,a]$
+
+#### Definition of commutator subgroup
+
+Let $G'$ be the subgroup of $G$ generated by all commutators of $G$.
+
+$$
+G'=\{[a_1,b_1][a_2,b_2]\ldots[a_n,b_n]\mid a_1,a_2,\ldots,a_n,b_1,b_2,\ldots,b_n\in G\}
+$$
+
+Then $G'$ is the subgroup of $G$.
+
+- Identity: $[e,e]=e$
+- Inverse: $([a_1,b_1],\ldots,[a_n,b_n])^{-1}=[b_n,a_n],\ldots,[b_1,a_1]$
+
+Some additional properties:
+
+- $G$ is abelian if and only if $G'=\{e\}$
+- $G'\trianglelefteq G$
+- $G/G'$ is abelian
+- If $N$ is a normal subgroup of $G$, and $G/N$ is abelian, then $G'\leq N$.
+
+> These implies that $G'$ is the smallest abelian normal subgroup of $G$.
+
+<details>
+<summary>Proof</summary>
+
+First we will show that $G'$ is normal
+
+Let $g\in G$ and $h=[a_1,b_1],\ldots,[a_n,b_n]\in G'$, then it is sufficient to show that $ghg^{-1}\in G'$.
+
+Consider arbitrary $[a,b]$
+
+$$
+\begin{aligned}
+g[a,b]g^{-1}&=ga_1 b_1 a_1^{-1}b_1^{-1}g^{-1}\\
+&=ga(g^{-1}g)b(g^{-1} g)a^{-1}(g^{-1}g)b^{-1}g^{-1}\\
+&=(gag^{-1})(gbg^{-1})(ga^{-1}g^{-1})(gb^{-1}g^{-1})\\
+&=(gag^{-1})(gbg^{-1})(gag^{-1})^{-1}(gbg^{-1})^{-1}\\
+&=[gag^{-1},gbg^{-1}]
+\end{aligned}
+$$
+
+---
+
+Them we will show that $G/G'$ is abelian.
+
+For all $a,b\in G$, we have $[a,b]\in G'$, so $[a,b]^{-1}\in G'$. Therefore $abG'=baG'\iff [b^{-1},a^{-1}]\in G'$. Therefore $G/G'$ is abelian.
+
+</details>

content/Math4302/_meta.js

@@ -21,4 +21,5 @@ export default {
     Math4302_L16: "Modern Algebra (Lecture 16)",
     Math4302_L17: "Modern Algebra (Lecture 17)",
     Math4302_L18: "Modern Algebra (Lecture 18)",
+    Math4302_L19: "Modern Algebra (Lecture 19)",
 }