Studying for a test I encountered this next question:
Let $S^1$ be the unit circle, $h:S^1\to S^1$ a continuous function which is homotopic to the constant function.
1) Prove that there is some $p\in S^1$ for which the angle between $p$ and $h(p)\in S^1$ is $90^{\circ}$.
2) (my own question) For what angles ($\theta\in[0,2\pi]$) does $p$ exist?
My thoughts:
I tried constructing some function which would help when using Brouwer fixed-point theorem, no luck so far.
There are continuous functions that does not meet the criteria, e.g $f(z)=ze^{i}$, but they are not homotopic to the constant function since they complete a full circle, so I assume that the composition $h\circ g:[0,1]\to S^1$ for $g:[0,2\pi]\to S^1, g(t)=e^{ti}$ does not complete a circle (am I right?) but I cannot see how I can put any boundary on the degree between a point and its image...
Brouwer's fixed point is a great tool. If $h\colon S^1 \rightarrow S^1$ is null-homopotic, then you can expand it to $H\colon B^2 \rightarrow S^1$ with $H\vert_{S^1} = h$. By Brouwer's fixed point THM, that $H$ has a fixed point $x\in B^2$ such that $H(x_0) = x_0\in S^1$, so $x_0$ is infact on the circle. We have, $$1=\langle H(x_0) ,x_0 \rangle=\langle h(x_0),x_0\rangle$$
If this helped you, then you should stop reading here and figure out how to get a negative sign for that function $x\mapsto \langle h(x),x\rangle$.
Consider $R\circ h$ with $R$ a rotation by $\pi$ (it is linear, thus continuous). This $R\circ h$ is obviouslly still null-homotpic, and we again obtain another $x^-$ such that $R\circ h(x^-) = x^-$. Inverting this, we have $$h(x^-)=R^{-1}\circ R\circ h(x^-)=R^{-1}(x^-)=-x^-$$ (i'm hope this isn't too confusing). Anyways: $$-1=\langle -x^- ,x^- \rangle =\langle h(x^-),x^-\rangle$$