4.3 KiB
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.
Recall that: Any map
g:S^1\to Yis nulhomotopic whenever it extends to a continuous mapG: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.
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.
Proof
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_*.