Math4202 Topology II (Lecture 23)

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.

Recall that: Any map $g:S^1\to Y$ is nulhomotopic whenever it extends to a continuous map $G:B^2\to Y$.

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 $B^2$.

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 $\tilde{f}(x)=f(Rx)$.

$$\begin{aligned} \tilde{f}(x) =f(Rx)&=(Rx)^n+a_{n-1}(Rx)^{n-1}+a_{n-2}(Rx)^{n-2}+\cdots+a_0\\ &=R^n\left(x^n+\frac{a_{n-1}}{R}x^{n-1}+\frac{a_{n-2}}{R^2}x^{n-2}+\cdots+\frac{a_0}{R^n}\right) \end{aligned}$$ $$\begin{aligned} \left\|\frac{a_{n-1}}{R}\right\|+\left\|\frac{a_{n-2}}{R^2}\right\|+\cdots+\left\|\frac{a_0}{R^n}\right\|&=\frac{1}{R}\|a_{n-1}\|+\frac{1}{R^2}\|a_{n-2}\|+\cdots+\frac{1}{R^n}\|a_0\|\\ &<\frac{1}{R}\left(\|a_{n-1}\|+\|a_{n-2}\|+\cdots+\|a_0\|\right)\\ &<\frac{1}{R}<1 \end{aligned}$$

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 $B^2_R$.

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 $h_*=k_*$.