Has the map $f:S^2 \to \mathbb{R} $ an antipodal point with the same image?

Question

The problem says that if $f:S^2 \to \mathbb{R} $ is continuous, then there exists a point $q\in S^2 $ such that $f(q)=f(-q)$.

I don't know how to prove it. Any help is welcome!

Answer 1

It's a direct consequence of the Borsuk-Ulam theorem. Apply it to

$$g:S^2\to\mathbb{R}^2$$ $$g(v)=\big(f(v), 0\big)$$

Answer 2

The B-U Theorem is hard to prove in dimension $2$ and greater, but simple in dimension $1$. Here's how to get your result using the dimension $1$ version.

Borsuk-Ulam, $1$-dim: If $g: S^1 \to \mathbb{R}$ is continuous then there is a point $x_0 \in S^1$ with $g(x_0) = g(-x_0)$.

Proof: Suppose not, so that $g(x) \neq g(-x)$ for all $x$. Define $h: S^1 \to \{\pm 1\}$ by the formula $$ h(x) = \frac{g(x)-g(-x)}{|g(x)-g(-x)|}. $$ This is well-defined because the bottom is never $0$. Moreover, $h$ is clearly continuous and onto, but this contradicts the connectedness of $S^1$. $\Box$

For your problem, embed $S^1$ into $S^2$ by the equator by the inclusion $i: S^1 \to S^2$. Now apply the above theorem to $g=f \circ i$ where $f$ is your map $f: S^2 \to \mathbb{R}$.

Answer 3

When I read Randall's elegant and short proof, I realized that we can easily prove the follwing more general result:

Let $X$ be a nonempty connected topological space with a continuous involution $\iota : X \to X$ (which means that $\iota \circ \iota = id_X$). Then for any continuous $g : X \to \mathbb{R}$ there exists a point $x \in X$ such that $g(\iota(x)) = g(x)$.

We may apply this to $X = S^n$ and $\iota(x) = -x$.

Define $h : X \to \mathbb{R}, h(x) = g(\iota(x)) - g(x)$. This is a continuous map such that $h(\iota(x)) = -h(x)$ for all $x \in X$. We conclude

$(1) \phantom{.}$ If $y \in h(X)$, then also $-y \in h(X)$.

Define $U = h(X) \cap (-\infty,0)$ and $V = h(X) \cap (0,\infty)$. These are open subsets of $h(X)$ and $(1)$ implies

$(2) \phantom{.}$ If $h(X) \ne \{ 0 \}$, then $U, V \ne \emptyset$.

Assume $0 \notin h(X)$. Then $h(X) = U \cup V$ and $U, V \ne \emptyset$ by $(2)$. Therefore $h(X)$ is not connected which contradicts the fact that continuous images of connected spaces are connected.

Therefore $0 \in h(X)$ which proves the assertion.