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.

Proof: part 1

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.

Proof: part 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}
\|\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.

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.

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