0

Prove that a map $f:S^1→S^1$ extends to the whole ball $B=\{|z|≤1\}$ if and only if $deg(f)=0$

again I can only do one direction

=>

$f:S^1→S^1$ is smooth, and $S^1 = \partial B$. Assume that $f:S^1→S^1$ extends to the whole ball$B$ then $deg(f)=0$ by the boundary theorem.

<=

Assume that $deg(f)=0$. The book tell me to extend $f$ to the annulus $A=\{\frac 12 \leq |z| \leq 1\}$ so that the inner circle $\{|z|= \frac 12\}$, the extended map is constant. And again I have no idea how this can be helpful to me.

Diane Vanderwaif
  • 4,292

1 Answers1

1

If $\deg(f) = 0$, then $f$ is null-homotopic; choose a homotopy $F_t:I \times S^1 \to S^1$ between $F_0 \equiv 1$ and $F_1 = f$. Define $\tilde f$ on $A$ by $\tilde f(r, \theta) = f_{2r - 1}(\theta)$ in polar coordinates. Since $\tilde f$ is constant on the circle $\{z:\, |z| = \frac{1}{2}\}$, it's trivial to extend it over the rest of the disk.

anomaly
  • 25,364
  • 5
  • 44
  • 85
  • so $\overline f$ is the extend map of $f$ . And the polar coordinate is $f=e^{i\theta}$? Can you tell me why $\overline f$ constant on the circle please? – Diane Vanderwaif Nov 16 '14 at 22:20
  • Just unwind the definition. – anomaly Nov 17 '14 at 04:25
  • one last question, it's trivial to extend over the rest of the disk if you know that it constant on the whole neighborhood of the inner circle. The question is is, how do you know this will hold? the map constant on the whole neighborhood of the inner circle – Diane Vanderwaif Nov 17 '14 at 23:31