# 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.

<details>
<summary>Proof</summary>

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 $h_*=k_*$.
</details>