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This question is unfortunately a 3-in-1 question, because Guillemin and Pollack's proof of Borsuk-Ulam relies on exercise 2.6.1 and exercise 2.6.2, the latter of which relies on exercise 2.4.8. I actually think I can do 2.6.1 and 2.6.2, but 2.4.8 is giving me quite a bit of trouble. Here is the statement:

Let $f:S^1\rightarrow S^1$ be any smooth map. Prove that there exists a smooth map $g:\mathbb{R}\rightarrow\mathbb{R}$ such that $f(\cos t, \sin t) = (\cos g(t), \sin g(t))$, and satisfying $g(2\pi) = g(0) + 2\pi q$ for some $q\in\mathbb{Z}$. [HINT: First define $g$ on $[0,2\pi]$, and show that $g(2\pi) = g(0) + 2\pi q$. Now extend $g$ by demanding $g(t + 2\pi) = g(t) + 2\pi q$.]

OK so... I'm fairly certain that $q$ will be the degree of $g$ as defined by the multiplication-by-$q$ map that $g_*$ is on homology. Here's how I really want to define this map: Let $g:[0,2\pi]\rightarrow \mathbb{R}$ be defined by $g(t) = \cos^{-1}(f_1(\cos t,\sin t)) + 2\pi k_t$ where $k_t$ is the number of times $\cos^{-1}(f_1(\cos t,\sin t))$ has appeared for $t' < t$ counting with a negative where $f$ is going clockwise at $t'$ and a positive where $f$ is going counterclockwise at $t'$. My only problem with this definition of a function is...well notice how long it took to even write! However, it will satisfy the requirements of the exercise, I think.

Any other ideas?

Tanner Strunk
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1 Answers1

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You are making things complicated.

You may as well consider continuous maps $F:\Bbb R\to S^1$ from the start. We claim that given $a_0$ with $F(0)=(\cos a_0,\sin a_0)$ there is a continuous $G$ lifting $F$ with $G(0)=a_0$, that is $F(t)=(\cos G(t),\sin G(t))$. As the hint points out, you can stick to defining $G$ on $[0,2\pi]$. As $F$ is uniformly continuous on $[0,2\pi]$ you can break up this interval into subintervals, $[x_j,x_{j+1}]$ where $0=x_0<x_1<\cdots<x_m=2\pi$ with the property that $|F(x)-F(x_j)|<1$ on $[x_j,x_{j+1}]$. The image of $F$ on $[x_j,x_{j+1}]$ is contained in a semicircle, and on each semicircle $t\mapsto(\cos t,\sin t)$ has a continuous inverse. So there is $G_j:[x_j,x_{j+1}]\to\Bbb R$ with $F(x)=(\cos G_j(x),\sin G_j(x))$ for $x\in[x_j,x_{j+1}]$. By adding suitable multiples of $2\pi$ to the $G_j$ you can align them at the endpoints, and paste together to make a continuous function $G$ on $[0,2\pi]$.

If you really want smoothness, note that each local inverse of $t\mapsto(\cos t,\sin t)$ is smooth.

Angina Seng
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