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Math4202 Topology II (Lecture 22)
Final reading, report, presentation
- Mar 30: Reading topic send email or discuss in OH.
- Apr 3: Finalize the plan.
- Apr 22,24: Last two lectures: 10 minutes to present.
- Final: type a short report, 2-5 pages.
Algebraic topology
Fundamental theorem of Algebra
For arbitrary polynomial f(z)=\sum_{i=0}^n a_i x^i. Are there roots in \mathbb{C}?
Consider f(z)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_0 is a continuous map from \mathbb{C}\to\mathbb{C}.
If f(z_0)=0, then z_0 is a root.
By contradiction, Then f:\mathbb{C}\to\mathbb{C}-\{0\}\cong \mathbb{R}^2-\{(0,0)\}.
Theorem for existence of n roots
A polynomial equation x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_0=0 of degree >0 with complex coefficients has at least one complex root.
There are n roots by induction.
Lemma
If g:S^1\to \mathbb{R}^2-\{(0,0)\} is the map g(z)=z^n, then g is not nulhomotopic. n\neq 0, n\in \mathbb{Z}.
Recall that we proved that
g(z)=zis not nulhomotopic.
Consider k:S^1\to S^1 by k(z)=z^n. k is continuous, k_*:\pi_1(S^1,1)\to \pi_1(S^1,1).
Where \pi_1(S^1,1)\cong \mathbb{Z}.
k_*(n)=nk_*(1).
Recall that the path in the loop p:I\to S^1 where p:t\mapsto e^{2\pi it}.
k_*(p)=[k(p(t))], where n=\tilde{k\circ p}(1).
k_* is injective.