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143 lines
4.4 KiB
Markdown
143 lines
4.4 KiB
Markdown
# Math4202 Topology II (Lecture 23)
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## Algebraic Topology
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### Fundamental Theorem of Algebra
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Recall the lemma $g:S^1\to \mathbb{R}-\{0\}$ is not nulhomotopic.
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$g=h\circ k$ 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)
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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)
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Therefore $g_*:\pi_1(S^1)\to \pi_1(\mathbb{R}-\{0\})$ is injective, therefore $g$ cannot be nulhomotopic. (nulhomotopic cannot be injective)
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#### Theorem
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Consider $x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_0=0$ of degree $>0$.
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<details>
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<summary>Proof: part 1</summary>
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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 $B^2$.
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We proceed by contradiction, suppose there is no root in $B^2$.
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Consider $f(x)=x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_0$.
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$f|_{B^2}$ is a continuous map from $B^2\to \mathbb{R}^2-\{0\}$.
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$f|_{S^1=\partial B^2}:S^1\to \mathbb{R}-\{0\}$ **is nulhomotopic**.
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> Recall that: Any map $g:S^1\to Y$ is nulhomotopic whenever it extends to a continuous map $G:B^2\to Y$.
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Construct a homotopy between $f|_{S^1}$ and $g$
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$$
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H(x,t):S^1\to \mathbb{R}-\{0\}\quad x^n+t(a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_0)
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$$
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Observer on $S^1$, $\|x^n\|=1,\forall n\in \mathbb{N}$.
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$$
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\begin{aligned}
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\|t(a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_0)\|&=t\|a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_0\|\\
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&\leq 1(\|a_{n-1}x^{n-1}\|+\|a_{n-2}x^{n-2}\|+\cdots+\|a_0\|)\\
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&=\|a_{n-1}\|+\|a_{n-2}\|+\cdots+\|a_0\|\\
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&<1
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\end{aligned}
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$$
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Therefore $H(s,t)>0\forall 0<t<1$. is a well-defined homotopy between $f|_{S^1}$ and $g$.
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Therefore $f_*=g_*$ is injective, $f$ is not nulhomotopic. This contradicts our previous assumption that $f$ is nulhomotopic.
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Therefore $f$ must have a root in $B^2$.
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</details>
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<details>
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<summary>Proof: part 2</summary>
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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)
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Consider $\tilde{f}(x)=f(Rx)$.
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$$
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\begin{aligned}
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\tilde{f}(x)
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=f(Rx)&=(Rx)^n+a_{n-1}(Rx)^{n-1}+a_{n-2}(Rx)^{n-2}+\cdots+a_0\\
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&=R^n\left(x^n+\frac{a_{n-1}}{R}x^{n-1}+\frac{a_{n-2}}{R^2}x^{n-2}+\cdots+\frac{a_0}{R^n}\right)
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\end{aligned}
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$$
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$$
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\begin{aligned}
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\left\|\frac{a_{n-1}}{R}\right\|+\left\|\frac{a_{n-2}}{R^2}\right\|+\cdots+\left\|\frac{a_0}{R^n}\right\|&=\frac{1}{R}\|a_{n-1}\|+\frac{1}{R^2}\|a_{n-2}\|+\cdots+\frac{1}{R^n}\|a_0\|\\
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&<\frac{1}{R}\left(\|a_{n-1}\|+\|a_{n-2}\|+\cdots+\|a_0\|\right)\\
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&<\frac{1}{R}<1
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\end{aligned}
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$$
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By Step 1, $\tilde{f}$ must have a root $z_0$ inside the unit disk.
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$f(Rz_0)=\tilde{f}(z_0)=0$.
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So $f$ has a root $Rz_0$ in $B^2_R$.
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</details>
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### Deformation Retracts and Homotopy Type
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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.
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For this section, we will show that $h_*$ is an isomorphism.
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#### Lemma for equality of homomorphism
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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.
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<details>
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<summary>Proof</summary>
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Let $H:X\times I\to Y$ be a homotopy from $h$ to $k$ such that
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$$
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H(x,0)=h(x), \qquad H(x,1)=k(x), \qquad H(x_0,t)=y_0 \text{ for all } t\in I.
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$$
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To show $h_*=k_*$, let $[f]\in \pi_1(X,x_0)$ be arbitrary, where
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$f:I\to X$ is a loop based at $x_0$, so $f(0)=f(1)=x_0$.
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Define
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$$
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F:I\times I\to Y,\qquad F(s,t)=H(f(s),t).
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$$
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Since $H$ and $f$ are continuous, $F$ is continuous. For each fixed $t\in I$, the map
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$$
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s\mapsto F(s,t)=H(f(s),t)
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$$
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is a loop based at $y_0$, because
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$$
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F(0,t)=H(f(0),t)=H(x_0,t)=y_0
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\quad\text{and}\quad
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F(1,t)=H(f(1),t)=H(x_0,t)=y_0.
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$$
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Thus $F$ is a based homotopy between the loops $h\circ f$ and $k\circ f$, since
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$$
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F(s,0)=H(f(s),0)=h(f(s))=(h\circ f)(s),
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$$
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and
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$$
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F(s,1)=H(f(s),1)=k(f(s))=(k\circ f)(s).
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$$
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Therefore $h\circ f$ and $k\circ f$ represent the same element of $\pi_1(Y,y_0)$, so
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$$
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[h\circ f]=[k\circ f].
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$$
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Hence
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$$
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h_*([f])=[h\circ f]=[k\circ f]=k_*([f]).
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$$
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Since $[f]$ was arbitrary, it follows that $h_*=k_*$.
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</details>
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