Let $f : S^2 \to \mathbb{R}$ be a continuous map such that $f(-x) = -f(x)$. Consider the set $Z = f^{-1}(0)$. Must $Z$ contain some path from some point to its antipode? Indeed, must $Z$ contain a continuous loop intersecting each "meridian", passing through antipodal points on antipodal meridians?


[I see this often asserted as a step in intuitive arguments for the Borsuk-Ulam theorem, but do not see why rigorously it must be so. Indeed, as a child, I once attempted precisely this argument for the Borsuk-Ulam theorem in a math camp, and was chided for asserting this unsupported lemma; thus it has stuck with me always. Certainly, $Z$ must contain a point on each meridian, but that these points must line up into a continuous path is not obvious to me.]


Update: My original question turned out to have already been answered at [How bogus is the glitzy proof of Borsuk-Ulam?](https://mathoverflow.net/questions/251921/how-bogus-is-the-glitzy-proof-of-borsuk-ulam), as well as by Loïc Teyssier in the same way below, but a new question with same motivation is: must Z contain two antipodal points in the same connected component? (If so, then the Borsuk-Ulam proof via this approach can still be salvaged; if not, we may write it off as a lost cause)



Must any continuous odd map from $\mathbb{S}^2$ to $\mathbb{R}$ have a path of zeros between antipodal points?
Let $K\subset \mathbb S^2$ be the compact set obtained by modifying the equator of $\mathbb S^2$ around two antipodal points so that it locally looks like the adherence of the graph of $t\neq0\mapsto \sin \frac{1}{t}$. You can make these modifications so that $K$ is symmetric for the antipodal involution. It is compact and not locally connected at the pair of points. In particular no two antipodal points in $K$ can be joined by a continuous path ranging in $K$.


The open set $\mathbb S^2\setminus K$ has two connected components, a "north" (say +) and a "south" (say -) domains, which are in antipodal involution. Let $f$ be the "distance to $K$" function multiplied by $\pm1$ according to the sign chosen for the north and south domains. Then $f$ is continuous, odd and $Z=K$.


Do you mean instead that some point of $Z$ should lie in the same connected component as its antipode?


