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?
The answer to the connected component version is yes, though the only proof I can come up with seems like more trouble than it would've been worth in math camp homework! Two steps:
- If there there are finitely many components of $S^2 \smallsetminus Z$, produce a finite-diameter bipartite tree on which the antipodal map on $S^2$ induces an involution fixing a point.
- Approximate any $f$ by functions satsifying step 1.
Most of the work seems to be a part of Step 1 that vaguely sounded like the Jordan curve theorem:
Key Fact. Let $X$ be a connected, compact, locally path-connected metric space with zero first homology. For an open subset $U$, inclusion induces a bijection between the connected components of $\partial U = \bar{U} \smallsetminus U$ and those of $X \smallsetminus U$.
Proof of Key Fact: Mayer-Vietoris and some point-set debris
Given $\epsilon > 0$, let $U_\epsilon$ be the set of points less than $\epsilon$ away from the closure $\bar{U}$ of $U$; and let $V_\epsilon$ be the set less than $\epsilon$ away from $X \smallsetminus U$. The Mayer-Vietoris sequence in reduced homology ends with
$$ H_1(X) \to \tilde{H}_0(U_\epsilon \cap V_\epsilon) \to \tilde{H}_0(U_\epsilon) \oplus \tilde{H}_0(V_\epsilon) \to \tilde{H}_0(X) , $$
which gives a bijection between the path components of $U_\epsilon \cap V_\epsilon$ and those of $V_\epsilon$. Pass to connected components of their closures using:
A connected open set $W$ in a locally path-connected space is path-connected.
Proof. Each point of $W$ is has a path-connected open neighborhood in $W$; so the path components form a partition of $W$ by open sets.
The closure of a connected open set $W$ is connected.
Proof. In a partition of $\bar{W}$ by clopen subsets, whichever one contains $W$ contains $\bar{W}$.
Then send $\epsilon$ to $0$. That is, since
$$ \begin{align*} X \smallsetminus U &= \bigcap_{\epsilon > 0} \bar{V}_\epsilon & &\text{and} & \partial U &= \bigcap_{\epsilon > 0} \overline{U_\epsilon \cap V_\epsilon} , \end{align*} $$
the Key Fact is proven if we can biject connected components of $X \smallsetminus U$ with nested sequences of components $C_n$ of $\bar{V}_{1/n}$ (and similarly for $\partial U$):
Lemma. In a compact space $X$, the intersection $C$ of nested closed connected sets $C_1 \supseteq C_2 \supseteq \cdots$ is connected.
Proof. Given disjoint open subsets $U$ and $V$ of $X$ that cover $C$, their union and the sequence $X \smallsetminus C_n$ form an open cover of $X$—which has a finite subcover, so some $C_n$ lies in $U \cup V$. Since $C_n$ is connected, it lies entirely in one of $U$ or $V$; and so does $C$.
Step 1: Turn connectivity data into a tree
Claim. Let $Z$ be the zero locus of a continuous odd $f: S^2 \to \mathbb{R}^2$. If $S^2 \smallsetminus Z$ has finitely many connected components, then the antipodal map on $S^2$ sends some component of $Z$ to itself.
Let $G$ be the graph where:
- Vertices are connected components of $Z$ and $S^2 \smallsetminus Z$, the sets of which we respectively denote $V_0$ and $V_{\neq 0}$.
- Two vertices are joined by an edge if their union is a connected set in $S^2$—equivalently, if $K \in V_0$ meets the closure of $U \in V_{\neq 0}$.
$G$ is bipartite with parts $V_0$ and $V_{\neq 0}$, and the assumption of Step 1 is that $V_{\neq 0}$ is finite; so—since $S^2$ is connected—$G$ is connected with finite diameter. By the Key Fact, every vertex in $V_{\neq 0}$ is a cut vertex, which makes $G$ a tree.
Every involution of a finite-diameter tree preserves either an edge or a vertex; and since the antipodal map can't exchange the parts of $G$ or fix any member of $V_{\neq 0}$, it has to fix exactly one vertex in $V_0$. This is the connected component of $Z$ we're after.
Step 2: Infinitely many components in $S^2 \smallsetminus Z$ is okay
Approximate $f$ by a sequence $f_n$ where $f_n$ is equal to $f$ on the connected components of $\{f \neq 0\}$ admitting an open ball of radius $\pi/n$ and zero everywhere else.
The area of an open ball of radius $\pi/n$ on $S^2$ is at least $4\pi/n^2$ (the area of a sphere of circumference $2\pi/n$). So each $f_n$ has no more than $n^2$ components to its nonzero locus; and running them through Step 1 yields a sequence of nested closed connected sets $C_1 \supseteq C_2 \supseteq \cdots$ preserved by the antipodal map.
Then the intersection of the sets $C_n$ is preserved by the antipodal map, in $Z$ (every component of $S^2 \smallsetminus Z$, being open, contains an open ball), and connected (by the Lemma before Step 1).