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content/Math4202/Math4202_L19.md

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+# Math4202 Topology II (Lecture 19)
+
+## Exam announcement
+
+Cover from first lecture to the fundamental group of circle.
+
+## Algebraic Topology
+
+### Retraction and fixed point
+
+#### Definition of retraction
+
+If $A\subseteq X$, a retraction of $X$ onto $A$ is a continuous map $r:X\to A$ such that $r|_A$ is the identity map of $A$.
+
+When such a retraction $r$ exists, $A$ is called a retract of $X$.
+
+<details>
+<summary>Example</summary>
+
+Identity map is a retraction of $X$ onto $X$.
+
+---
+
+$X=\mathbb{R}^2$, $A=\{0\}$, the constant map that maps all points to $(0,0)$ is a retraction of $X$ onto $A$.
+
+This can be generalized to any topological space, take $A$ as any one point set in $X$.
+
+---
+
+Let $X=\mathbb{R}^2$, $A=\mathbb{R}$, the projection map that maps all points to the first coordinate is a retraction of $X$ onto $A$.
+
+---
+
+> Can we retract $\mathbb{R}^2$ to a circle?
+
+Let $\mathbb{R}^2\to S^1$
+
+This can be done in punctured plane. $\mathbb{R}^2\setminus\{0\}\to S^1$. by $\vec{x}\mapsto \vec{x}/\|x\|$.
+
+But
+
+</details>
+
+#### Lemma for retraction
+
+If $A$ is a retract of $X$, the homomorphism of fundamental groups induced by the inclusion map $j:A\to X$, with induced $j_*:\pi_1(A,x_0)\to \pi_1(X,x_0)$ is injective.
+
+<details>
+<summary>Proof</summary>
+
+Let $r:X\to A$ be a retraction. Consider $j:A\to X, r:X\to A$. Then $r\circ j(a)=r(a)=a$. Therefore $r\circ j=Id_A$.
+
+Then $r_*\circ j_*=Id_{\pi_1(A,x_0)}$.
+
+$\forall f\in \ker j_*$, $j_*f=0$. $r_*\circ j_*f=Id_{f}=f$, therefore $f=0$.
+
+So $\ker j_*=\{0\}$.
+
+So it is injective.
+</details>
+
+Consider the $\mathbb{R}^2\to S^1$ example, if such retraction exists, $j_*:\pi_1(S^1,x_0)\to \pi_1(\mathbb{R}^2,x_0)$ is injective. But the fundamental group of circle is $\mathbb{Z}$ whereas the fundamental group of plane is $1$. That cannot be injective.
+
+#### Corollary for lemma of retraction
+
+There is no retraction from $\mathbb{R}^2$, $B_1(0)\subseteq \mathbb{R}^2$ (unit ball in $\mathbb{R}^2$), to $S^1$.
+
+#### Lemma
+
+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 from $B_1(0)\to X$.
+- $h_*$ 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).
+

content/Math4202/_meta.js

@@ -20,4 +20,6 @@ export default {
     Math4202_L15: "Topology II (Lecture 15)",
     Math4202_L16: "Topology II (Lecture 16)",
     Math4202_L17: "Topology II (Lecture 17)",
+    Math4202_L18: "Topology II (Lecture 18)",
+    Math4202_L19: "Topology II (Lecture 19)",
 }

content/Math4302/Math4302_L18.md

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+# Math4302 Modern Algebra (Lecture 18)
+
+## Groups
+
+### Factor group
+
+Suppose $G$ is a group, and $H\trianglelefteq G$, then $G/H$ is a group.
+
+Recall from last lecture, if $\phi:G\to G'$ is a homomorphism, then $G/\ker(\phi)\simeq \phi(G)\leq G'$.
+
+<details>
+<summary>Example (continue from last lecture)</summary>
+
+$\mathbb{Z}\times\mathbb{Z}/\langle (1,1)\rangle\simeq \mathbb{Z}$
+
+Take $\phi(a,b)=a-b$, this is a surjective homomorphism from $\mathbb{Z}\times\mathbb{Z}/\langle (1,1)\rangle$ to $\mathbb{Z}$
+
+---
+
+$\mathbb{Z}\times\mathbb{Z}/\langle (2,1)\rangle\simeq \mathbb{Z}$
+
+where $\langle (2,1)\rangle=\{(2b,b)|b\in \mathbb{Z}\}$
+
+Take $\phi(a,b)=a-2b$, this is a surjective homomorphism from $\mathbb{Z}\times\mathbb{Z}/\langle (2,1)\rangle$ to $\mathbb{Z}$
+
+---
+
+$\mathbb{Z}\times\mathbb{Z}/\langle (2,2)\rangle$
+
+This should also be a finitely generated abelian group. ($\mathbb{Z}_2\times \mathbb{Z}$ actually)
+
+Take $\phi(a,b)=(a\mod 2,a-b)$
+
+---
+
+More generally, for $\mathbb{Z}\times \mathbb{Z}/\langle (a,b)\rangle$.
+
+This should be $\mathbb{Z}\times \mathbb{Z}_{\operatorname{gcd}(a,b)}$
+
+Try to do section by gcd.
+
+</details>
+
+> - If $G$ is abelian, $N\leq G$, then $G/N$ is abelian.
+> - If $G$ is finitely generated and $N\trianglelefteq G$, then $G/N$ is finitely generated.
+
+#### Definition of simple group
+
+$G$ is simple if $G$ has no proper ($H\neq G,\{e\}$), normal subgroup.
+
+> [!TIP]
+> 
+> In general $S_n$ is not simple, consider the normal subgroup $A_n$.
+
+<details>
+<summary>Example of some natural normal subgroups</summary>
+
+If $\phi:G\to G'$ is a homomorphism, then $\ker(\phi)\trianglelefteq G$.
+
+---
+
+The **center** of $G$: $Z(G)=\{a\in G|ag=ga\text{ for all }g\in G\}$
+
+$Z(G)\trianglelefteq G$.
+
+- $e\in Z(G)$.
+- $a,b\in Z(G)\implies abg=gab\implies ab\in Z(G)$.
+- $a\in Z(G)\implies ag=ga\implies a^{-1}\in Z(G)$.
+- If $g\in G, h\in Z(G)$, then $ghg^{-1}\in Z(G)$ since $ghg^{-1}=gg^{-1}h=h$.
+
+$Z(S_3)=\{e\}$, all the transpositions are not commutative, so $Z(S_3)=\{e\}$.
+
+$Z(GL_n(\mathbb{R}))$? continue on friday.
+
+</details>

content/Math4302/_meta.js

@@ -20,4 +20,5 @@ export default {
     Math4302_L15: "Modern Algebra (Lecture 15)",
     Math4302_L16: "Modern Algebra (Lecture 16)",
     Math4302_L17: "Modern Algebra (Lecture 17)",
+    Math4302_L18: "Modern Algebra (Lecture 18)",
 }