show/hide this revision's text 2 minor correction and clarification

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 $\gamma:\mathbb S^1\to\mathbb R^2$. Choose orientations so that the left half of the , namely $\infty$ is followed clockwise and the right half is followed counter-clockwise. I assume that $ \gamma$ is contained in a unit square and its two tangent lines at the center are the bisectors of the coordinate angles\gamma(t) = (\sin t,\tfrac12\sin2t) $$ (here $\mathbb S^1=\mathbb R/2\pi\mathbb Z$). Then 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 unit square 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 $D^-$.
show/hide this revision's text 1

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 $\gamma:\mathbb S^1\to\mathbb R^2$. Choose orientations so that the left half of the $\infty$ is followed clockwise and the right half is followed counter-clockwise. I assume that $\gamma$ is contained in a unit square and its two tangent lines at the center are the bisectors of the coordinate angles. Then 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 unit square neighborhoods.

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 $D^-$.