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content/Math4202/Math4202_L23.md

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+# Math4202 Topology II (Lecture 23)
+
+## Algebraic Topology
+
+### Fundamental Theorem of Algebra
+
+Recall the lemma $g:S^1\to \mathbb{R}-\{0\}$ is not nulhomotopic.
+
+$g=h\circ k$ where $k:S^1\to S^1$ by $z\mapsto z^n$, $k_*:\pi_1(S^1)\to \pi_1(S^1)$ is injective. (consider the multiplication of integer is injective)
+
+and $h:S^1\to \mathbb{R}-\{0\}$ where $z\mapsto z$. $h_*:\pi_1(S^1)\to \pi_1(\mathbb{R}-\{0\})$ is injective. (inclusion map is injective)
+
+Therefore $g_*:\pi_1(S^1)\to \pi_1(\mathbb{R}-\{0\})$ is injective, therefore $g$ cannot be nulhomotopic. (nulhomotopic cannot be injective)
+
+#### Theorem
+
+Consider $x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_0=0$ of degree $>0$.
+
+<details>
+<summary>Proof: part 1</summary>
+
+Step 1: if $|a_{n-1}|+|a_{n-2}|+\cdots+|a_0|<1$, then $x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_0=0$ has a root in the unit disk $B^2$.
+
+We proceed by contradiction, suppose there is no root in $B^2$.
+
+Consider $f(x)=x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_0$.
+
+$f|_{B^2}$ is a continuous map from $B^2\to \mathbb{R}^2-\{0\}$.
+
+$f|_{S^1=\partial B^2}:S^1\to \mathbb{R}-\{0\}$ **is nulhomotopic**.
+
+Construct a homotopy between $f|_{S^1}$ and $g$
+
+$$
+H(x,t):S^1\to \mathbb{R}-\{0\}\quad x^n+t(a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_0)
+$$
+
+Observer on $S^1$, $\|x^n\|=1,\forall n\in \mathbb{N}$.
+
+$$
+\begin{aligned}
+\|t(a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_0)\|&=t\|a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_0\|\\
+&\leq 1(\|a_{n-1}x^{n-1}\|+\|a_{n-2}x^{n-2}\|+\cdots+\|a_0\|)\\
+&=\|a_{n-1}\|+\|a_{n-2}\|+\cdots+\|a_0\|\\
+&<1
+\end{aligned}
+$$
+
+Therefore $H(s,t)>0\forall 0<t<1$. is a well-defined homotopy between $f|_{S^1}$ and $g$.
+
+Therefore $f_*=g_*$ is injective, $f$ is not nulhomotopic. This contradicts our previous assumption that $f$ is nulhomotopic.
+
+Therefore $f$ must have a root in $B^2$.
+
+</details>
+
+<details>
+<summary>Proof: part 2</summary>
+
+If \|a_{n-1}\|+\|a_{n-2}\|+\cdots+\|a_0\|< R$ has a root in the disk $B^2_R$. (and $R\geq 1$, otherwise follows part 1)
+
+Consider $\tilde{f}(x)=f(Rx)$. 
+
+$$
+\begin{aligned}
+\tilde{f}(x)
+=f(Rx)&=(Rx)^n+a_{n-1}(Rx)^{n-1}+a_{n-2}(Rx)^{n-2}+\cdots+a_0\\
+&=R^n\left(x^n+\frac{a_{n-1}}{R}x^{n-1}+\frac{a_{n-2}}{R^2}x^{n-2}+\cdots+\frac{a_0}{R^n}\right)
+\end{aligned}
+$$
+
+$$
+\begin{aligned}
+\|\frac{a_{n-1}}{R}\|+\|\frac{a_{n-2}}{R^2}\|+\cdots+\|\frac{a_0}{R^n}\|&=\frac{1}{R}\|a_{n-1}\|+\frac{1}{R^2}\|a_{n-2}\|+\cdots+\frac{1}{R^n}\|a_0\|\\
+&<\frac{1}{R}\left(\|a_{n-1}\|+\|a_{n-2}\|+\cdots+\|a_0\|\right)\\
+&<\frac{1}{R}<1
+\end{aligned}
+$$
+
+By Step 1, $\tilde{f}$ must have a root $z_0$ inside the unit disk.
+
+$f(Rz_0)=\tilde{f}(z_0)=0$.
+
+So $f$ has a root $Rz_0$ in $B^2_R$.
+
+</details>
+
+### Deformation Retracts and Homotopy Type
+
+Recall previous section, $h:S^1\to \mathbb{R}-\{0\}$ gives $h_*:\pi_1(S^1,1)\to \pi_1(\mathbb{R}-\{0\},0)$ is injective.
+
+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.