Suppose $f: S^2 \rightarrow {\bf R}^2$ is continuous; let $A$ be the set of points $u \in S^2$ such that $f(u)-f(-u) \in {\bf R} \times \{0\}$ (where $-u$ denotes the antipode of $u$). Given $u,-u \in A$, must there exist a path in $A$ joining $u$ to $-u$?
The appealing argument presented at https://www.youtube.com/watch?v=csInNn6pfT4 (which I have also seen elsewhere), which purports to prove the existence of a $u \in S^2$ with $f(u)-f(-u) = (0,0)$, hinges on this unstated assumption.
Can anyone devise an $f$ for which $A$ is something like a topologist's sine curve?
ADDED LATER: I like Ilya Bogdanov's example, and I wonder whether we can improve it. Is there an $f$ such that no point in $A$ can be joined to its antipode by a continuous curve?
Best Answer
Let $S^2$ be the unit sphere in $\mathbb R^3$, $S^2_+=\{(x,y,z)\in S^2\mid z\geq0\}$ the upper hemisphere, $S^2_-$ the lower hemisphere and $B^2$ the closed unit disk in $\mathbb R^2$. Define $p:S^2\to B^2$ by $p(x,y,z)=(x,y)$. Then the restrictions of $p_+:S^2_+\to B^2$ and $p_-:S^2_-\to B^2$ of $p$ are homeomorphisms. Define a copy of the topologist's sine wave in $B^2$ by $$W=\overline{\{(x,y)\in B^2\mid x\neq 0\text{ and } y = \tfrac12\sin\tfrac{\pi}x\}}.$$ Then, $K=p_+^{-1}(W)\cup p_-^{-1}(W)$ is a closed subset of $S^2$ with the following properties:
Now consider $f:S^2\to\mathbb R^2$ defined by $$f(x,y,z)=(0,\pm d((x,y,z),K)),$$ where $d$ is e.g. the Euclidean metric on the sphere and the sign is $+$ for $(x,y,z)\in U_+$ and $-$ for $(x,y,z)\in U_-$.
This function $f$ has all the desired properties: it is continuous and the set of points $u=(x,y,z)$ such that $f(u)-f(-u)\in\mathbb R\times\{0\}$ is precisely $K$, because for any $u\in S^2\setminus K$, the second components of $f(u)$ and $f(-u)$ have different signs.