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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?, 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)

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  • $\begingroup$ For what it's worth, I now find the winding number proof of Borsuk-Ulam much more intuitive and illuminating (given odd map from sphere to R^2, its winding number around 0 as its input traces around the equator must be odd and thus nonzero. But winding numbers around boundaries add as regions combine, and also vanish in the vicinity of non-zeros of the map. Thus, by repeated bisection of input loops, we can constructively zoom in on some point which is a zero of the map.). I'm just curious whether this other approach that everyone goes for can indeed be fleshed out to hold up or not. $\endgroup$ – Sridhar Ramesh Oct 29 '17 at 23:55
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    $\begingroup$ I've just noticed essentially identical question: mathoverflow.net/questions/251921/… $\endgroup$ – Sridhar Ramesh Oct 30 '17 at 1:58
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    $\begingroup$ I've updated this question, and am now very interested in whether Z must contain antipodal points in the same connected component. $\endgroup$ – Sridhar Ramesh Oct 30 '17 at 2:44
  • $\begingroup$ As noted, I'm still very interested in the connected component version of this question… Not sure if it's best to keep bumping this question or to make a new one. $\endgroup$ – Sridhar Ramesh Nov 5 '17 at 23:30
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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?

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  • $\begingroup$ This f isn't odd, though, since distances are always nonnegative. Can this f be odd-ized? $\endgroup$ – Sridhar Ramesh Oct 29 '17 at 23:37
  • $\begingroup$ Ah... you're right of course. I'll edit the answer. $\endgroup$ – Loïc Teyssier Oct 29 '17 at 23:40
  • $\begingroup$ Great, thanks! And, sure, I'd also be interested in the connected component version of this, since I suppose Z having two antipodal points in same connected component still allows us to conclude that any other odd function from S^2 to R must have a zero in the same Z as well, and thus obtain Borsuk-Ulam. $\endgroup$ – Sridhar Ramesh Oct 29 '17 at 23:59
  • $\begingroup$ The zero set can be made to have no simple curves. $\endgroup$ – Anton Petrunin Oct 30 '17 at 1:49
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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:

  1. 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.
  2. 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:

  1. 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.

  2. 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).

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  • $\begingroup$ Thank you so much for resolving this! I wish I could accept two answers to this question. $\endgroup$ – Sridhar Ramesh Nov 13 '17 at 0:21

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