Generalization of Borsuk-Ulam

Let $n$ be a positive intger. Is the following true? For continuous maps $f: \mathbb S^n \rightarrow \mathbb S^n$ and $g: \mathbb S^n \rightarrow \mathbb R^n$, there exists a point $x \in \mathbb S^n$ such that $g(x) = g(f(x))$.

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OK, so I've been thinking about your problem a little bit, and I'm still a little bit unclear. Why do you call this statement a generalization of Borsuk-Ulam? – Thierry Zell Apr 9 '11 at 17:05
@Thierry: If f is the antipodal map of the sphere, then this is exactly the Borsuk-Ulam Theorem. – Bill Kronholm Apr 9 '11 at 17:12
Thanks Bill, I guess I'm not very awake this morning! – Thierry Zell Apr 9 '11 at 17:16

This is false for all $n \geq 2$, but true for $n=1$.

A map $g \colon X \to Y$ of topological space is said to be free if there is a map $f \colon X \to X$ such that $g(x) \neq g(f(x))$ for all $x\in X$. This definition appears to be due to Hopf, and was given in

H. Hopf, Freie Überdeckungen und freie Abbildungen, Fund. Math. 28 (1937), 33--57.

In that same paper, Hopf apparently proved that:

1. There is a free map $g \colon \mathbb S^n \to \mathbb R^n$ for every $n \geq 3$.
2. There is no free map from $\mathbb S^n$ into $\mathbb R$ if $n \geq 1$.

A generalization was obtained by Pannwitz in

E. Pannwitz, Eine freie Abbildung der n-dimensionalen Sphäre in die Ebene. Math. Nachr. 7 (1952), 183–-185.

According to the MR, she proved that, for every $n \geq 0$, there exists a free map $g \colon \mathbb S^n \to \mathbb R ^2$.

Unfortunately, I'm having trouble accessing the papers, so I can't provide any more details.

Update. Still no luck getting Pannwitz's paper. However, I have managed to find the following related paper

C. Biasi, D. de Mattos, E. dos Santos, Applications of the non-standard version of the Borsuk--Ulam theorem, JP J. Geom. Topol. 9 (2009), no. 3, 273--284.

(available online here), which mentions the above papers of Hopf and Pannwitz in its Introduction. More interesting however is Theorem 2.1, which I believe can be applied to Sergei's construction to yield Pannwitz's result for $n\geq 2$. This would show that the question in the OP has a negative answer if $n\geq 2$.

It remains to show that the answer is "yes" if $n=1$. (As Harry notes, the answer is also "yes" if $n=0$.) The following is Hopf's argument. Let $g \colon \mathbb S^1 \to \mathbb R$ and $f \colon \mathbb S^1 \to \mathbb S^1$ be given. Since $\mathbb S^1$ is compact, $g$ attains its maximum, resp. minimum, at some $x$, resp. $y$, in $\mathbb S^1$. In particular, $g(x) \geq g(f(x))$ and $g(y) \leq g(f(y))$. The intermediate value theorem then asserts that there is a $z \in \mathbb S^1$ for which $g(z) = g(f(z))$.

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It's also trivially true if n=0. :) – Harry Altman Apr 10 '11 at 15:31

This is not true for $n=2$.

Represent $\mathbb S^2$ as the cylinder $\mathbb S^1\times[-1,1]$ with two discs $D_+$ and $D_-$ attached to the boundary components $\mathbb S^1\times\{1\}$ and $\mathbb S^1\times\{-1\}$, resp. Denote by $\mathbb S^2_+$ and $\mathbb S^2_-$ the "positive" and "negative" hemispheres: $\mathbb S^2_+=(\mathbb S^1\times[0,1])\cup D_+$ and $\mathbb S^2_-$ is the opposite.

First we define $g:\mathbb S^2\to\mathbb R^2$. Consider a smooth $\infty$-shaped loop $\gamma:\mathbb S^1\to\mathbb R^2$, namely $$\gamma(t) = (\sin t,\tfrac12\sin2t)$$ (here $\mathbb S^1=\mathbb R/2\pi\mathbb Z$). Note that the velocity of $\gamma$ is separated away from the vertical vector $e_2=(0,1)$, namely $\angle(\dot\gamma(t),e_2)\ge\pi/4$ for all $t\in\mathbb S^1$. Choose $\varepsilon>0$ so small that $\angle (\gamma(t+\varepsilon)-\gamma(t),e_2)> \pi/5$ for all $t$.

For $x=(t,s)$ from the cylinder, define $$\begin{cases} g(x) =\gamma(t)+1000s\cdot\overrightarrow{(1,10)} , &\qquad s\ge 0 \cr g(x) =\gamma(t)+1000|s|\cdot\overrightarrow{(-1,10)} , &\qquad s\le 0 . \end{cases}$$ (its image of the cylinder consists of two almost vertical strips above the $\infty$-figure). Observe that the image of each boundary component $\mathbb S^1\times\{\pm 1\}$ is separated away from the image of the other half of the cylinder (by distance at least 100). Extend $g$ to $D_+$ and $D_-$ so as to fill these boundary components within their neighborhoods of radius 2. Then $g(D_+)\cap g(\mathbb S^2_-)=\emptyset$ and $g(D_-)\cap g(\mathbb S^2_+)=\emptyset$.

Since $\gamma(t+\varepsilon)-\gamma(t)$ never forms a small angle with $e_2$, the construction guarantees that $g(t,s)\ne\gamma(t+\varepsilon)$ for all $t\in\mathbb S^1$, $s\in[-1,1]$.

Now we define $f:\mathbb S^n\to\mathbb S^n$. For $x=(t,s)$ from the cylinder, define $f(x)=(t+\varepsilon,0)$, so the cylinder is projected to its equator and slightly rotated. Extend $f$ to $D_+$ and $D_-$ so that $f(D_+)\subset \mathbb S^2_-$ and $f(D^-)\subset \mathbb S^2_+$.

For these $f$ and $g$, we have $g(x)\ne g(f(x))$ for all $x\in\mathbb S^2$. Indeed, if $x=(t,s)\in\mathbb S^1\times[-1,1]$, then $f(g(x))=\gamma(t+\varepsilon)\ne g(x)$ as noted above. For $x\in D^+$ this follows from the fact that $g(D^+)\cap g(f(D^+))=\emptyset$, and similarly for $D^-$.

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Minor note, you seem to have clockwise and counterclockwise backwards. – Harry Altman Apr 10 '11 at 15:10
Thanks, corrected this by making the example more explicit. – Sergei Ivanov Apr 10 '11 at 22:05

On the other hand, it is known to be true if f is an involution, i.e. f(f(x))=x. This was proved in a paper by C.T. Yang in Annals of Mathematics in 1954.

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You mean this paper? jstor.org/stable/1969632 – Erel Segal-Halevi May 18 '15 at 9:22