2
$\begingroup$

The $2$-dimensional Borsuk-Ulam theorem states that if $f:\mathbb{S}^2\to \mathbb{R}^2$ is a continuous function, then there is some $x\in \mathbb{S}^2$ such that $f(x)=f(-x)$.

The proof relies on a previous result which states that there is no continuous map $f: \mathbb{S}^2\to\mathbb{S}^1$ such that $f(-x)=-f(x)$ for all $x \in \mathbb{S}^2$, and there is a step in this proof I don't manage to understand.

Take a path $\gamma$ that joins $x_0=(1,0,0)$ with $(-1,0,0)$ on $\mathbb{S}^2$, and take $\mu = f \circ \gamma$, which is a path joining $f(x_0)$ with $f(-x_0)$. Let $\psi: \mathbb{S}^1 \to \mathbb{S}^1$ be the covering map $$\psi: e^{2\pi i t} = e^{4\pi i t}$$ Then, $\psi \circ \mu$ is a loop that doesn't lift to a loop by the covering map $\psi$, which implies that $[\psi \circ \mu] \neq 0$ (where $[\sigma]$ denotes an element in fundamental group).

I don't understand the reasoning on the last sentence. I believe that the step $[\psi \circ \mu] \neq 0$ is justified by the fact that the lifting is not a loop, but that last part is what I can't prove.

Why isn't $\psi \circ \mu$ a loop?

Thank you in advance.

$\endgroup$

1 Answer 1

3
$\begingroup$

Missing from the section of proof you have included is the initial assumption that $f\colon S^2\to S^1$ satisfies $f(-x)=-f(x)$ for all $x\in S^2$.

Under that assumption, we have \begin{align} \mu(0) &= f(\gamma(0)) \\ &= f(1,0,0)\\ &= -f(-1,0,0)\\ &=-f(\gamma(1))\\ &=-\mu(1) \end{align}

In particular, $\mu(0)\ne\mu(1)$, so $\mu$ is not a loop.

However, since the covering map $\phi$ sends antipodal points to the same point ($e^{4\pi i(t + \pi/2)} = e ^{4\pi i} e^{2\pi i}=e^{4\pi i}$), $\psi(\mu(0))=\psi(\mu(1))$, so $\psi$ is a loop.

By the uniqueness part of the path lifting property (a special case of the Homotopy lifting property), $\mu$ is the unique lift of $\phi\circ\mu$ along $\phi$, and so $\phi\circ\mu$ admits no lift that is a loop.

$\endgroup$
1
  • $\begingroup$ Perfect! I need to work through the lifting property some more I see. Thank you very much! $\endgroup$
    – user313212
    Oct 19, 2017 at 15:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .