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Yashar Mermarian writes here that the answer is yes. And the argument he gives is pretty much the same as the one you already started.

Taking $n=k$ and $\epsilon=\pi/2$ Gromov's waist inequality gives a point $z\in \mathbb{R}^n$ such that $$vol(U_{\pi/2}(S_Z))\geq vol(U_{\pi/2}(S^0))=vol(U_{\pi/2}(\{(1,0,\dots,0),(-1,0,\dots,0)\})=vol(S^n).$$ It follows that $U_{\pi/2}(S_Z)$ has to have the same volume as the sphere. This can only be the case when $S_Z$ contains at least $2$ elements, since the volume of $U_\epsilon(S_Z)$ would be to small otherwise. If it contains $2$ elements and not more, these $2$ elements would have to be antipodal, otherwise the volume of $U_\epsilon(S_Z)$ would be to small again. Although Memarian writes there "is no choice" for $S_Z$ but to "pass through antipodal points", I don't really see how one can rule out that $S_Z$ might consist of more than two points. I hope someone can make this clear.

To summarize this gives the following version of a Bursuk Ulam type result:

If $f:\; S^n\to\mathbb{R}^n$ is continuous, then $f$ maps two antipodal points in $S^n$ to the same point in $\mathbb{R}^n$ or $f$ maps (at least) three points to the same point in $\mathbb{R}^n$.

$U_{\pi/2}$ for two points which are not antipodal.

In the original paper Gromov is more careful. He writes:

If $k=n$, and $\operatorname{card}(f^{-1}(z))\leq 2$, $z\in\mathbb{R}^n$, then [the waist of the sphere theorem] applied to $\pi/2$ amounts to the Borsuk–Ulam theorem: some level $f^{-1}(z)$ of $f\;:S^n\to \mathbb{R}^k$ equals a pair of opposite points.

Of course $f^{-1}(z)$ is just a different notation of $S_Z$. So the important detail is: "and $\operatorname{card}(f^{-1}(z))\leq 2$".

To summarize the answer is: Gromov's Waist Inequality almost implies Borsuk-Ulam.

Yashar Mermarian writes here that the answer is yes. And the argument he gives is pretty much the same as the one you already started.

Taking $n=k$ and $\epsilon=\pi/2$ Gromov's waist inequality gives a point $z\in \mathbb{R}^n$ such that $$vol(U_{\pi/2}(S_Z))\geq vol(U_{\pi/2}(S^0))=vol(U_{\pi/2}(\{(1,0,\dots,0),(-1,0,\dots,0)\})=vol(S^n).$$ It follows that $U_{\pi/2}(S_Z)$ has to have the same volume as the sphere. This can only be the case when $S_Z$ contains at least $2$ elements, since the volume of $U_\epsilon(S_Z)$ would be to small otherwise. If it contains $2$ elements and not more, these $2$ elements would have to be antipodal, otherwise the volume of $U_\epsilon(S_Z)$ would be to small again. Although Memarian writes there "is no choice" for $S_Z$ but to "pass through antipodal points", I don't really see how one can rule out that $S_Z$ might consist of more than two points.

This gives the following (weaker) version of a Bursuk Ulam type result:

If $f:\; S^n\to\mathbb{R}^n$ is continuous, then $f$ maps two antipodal points in $S^n$ to the same point in $\mathbb{R}^n$ or $f$ maps (at least) three points to the same point in $\mathbb{R}^n$.

$U_{\pi/2}$ for two points which are not antipodal.

In the original paper Gromov is more careful. He writes:

If $k=n$, and $\operatorname{card}(f^{-1}(z))\leq 2$, $z\in\mathbb{R}^n$, then [the waist of the sphere theorem] applied to $\pi/2$ amounts to the Borsuk–Ulam theorem: some level $f^{-1}(z)$ of $f\;:S^n\to \mathbb{R}^k$ equals a pair of opposite points.

Of course $f^{-1}(z)$ is just a different notation of $S_Z$. So the important detail is: "and $\operatorname{card}(f^{-1}(z))\leq 2$".

In general $\operatorname{card}(f^{-1}(z))\leq 2$ is not true for arbitrary maps. And even if you perturbe $f$ a little bit this condition won't be fullfilled. Of course any two maps from $S^n$ to $\mathbb{R}^n$ are homotopic, so in particular $f$ will be homotopic to a map with $\operatorname{card}(f^{-1}(z))\leq 2$ for all $z$, but this doesn't really give you anything.

To summarize the answer is: Gromov's Waist Inequality almost implies Borsuk-Ulam.

Yashar Mermarian writes here that the answer is yes. And the argument he gives is pretty much the same as the one you already started.

Taking $n=k$ and $\epsilon=\pi/2$ Gromov's waist inequality gives a point $z\in \mathbb{R}^n$ such that $$vol(U_{\pi/2}(S_Z))\geq vol(U_{\pi/2}(S^0))=vol(U_{\pi/2}(\{(1,0,\dots,0),(-1,0,\dots,0)\})=vol(S^n).$$ It follows that $U_{\pi/2}(S_Z)$ has to have the same volume as the sphere. This can only be the case when $S_Z$ contains at least $2$ elements, since the volume of $U_\epsilon(S_Z)$ would be to small otherwise. If it contains $2$ elements and not more, these $2$ elements would have to be antipodal, otherwise the volume of $U_\epsilon(S_Z)$ would be to small again. Although Memarian writes there "is no choice" for $S_Z$ but to "pass through antipodal points", I don't really see how one can rule out that $S_Z$ might consist of more than two points. I hope someone can make this clear.

To summarize this gives the following version of a Bursuk Ulam type result:

If $f:\; S^n\to\mathbb{R}^n$ is continuous, then $f$ maps two antipodal points in $S^n$ to the same point in $\mathbb{R}^n$ or $f$ maps (at least) three points to the same point in $\mathbb{R}^n$.

$U_{\pi/2}$ for two points which are not antipodal.

Yashar Mermarian writes here that the answer is yes. And the argument he gives is pretty much the same as the one you already started.

Taking $n=k$ and $\epsilon=\pi/2$ Gromov's waist inequality gives a point $z\in \mathbb{R}^n$ such that $$vol(U_{\pi/2}(S_Z))\geq vol(U_{\pi/2}(S^0))=vol(U_{\pi/2}(\{(1,0,\dots,0),(-1,0,\dots,0)\})=vol(S^n).$$ It follows that $U_{\pi/2}(S_Z)$ has to have the same volume as the sphere. This can only be the case when $S_Z$ contains at least $2$ elements, since the volume of $U_\epsilon(S_Z)$ would be to small otherwise. If it contains $2$ elements and not more, these $2$ elements would have to be antipodal, otherwise the volume of $U_\epsilon(S_Z)$ would be to small again. Although Memarian writes there "is no choice" for $S_Z$ but to "pass through antipodal points", I don't really see how one can rule out that $S_Z$ might consist of more than two points. I hope someone can make this clear.

To summarize this gives the following version of a Bursuk Ulam type result:

If $f:\; S^n\to\mathbb{R}^n$ is continuous, then $f$ maps two antipodal points in $S^n$ to the same point in $\mathbb{R}^n$ or $f$ maps (at least) three points to the same point in $\mathbb{R}^n$.

$U_{\pi/2}$ for two points which are not antipodal.

In the original paper Gromov is more careful. He writes:

If $k=n$, and $\operatorname{card}(f^{-1}(z))\leq 2$, $z\in\mathbb{R}^n$, then [the waist of the sphere theorem] applied to $\pi/2$ amounts to the Borsuk–Ulam theorem: some level $f^{-1}(z)$ of $f\;:S^n\to \mathbb{R}^k$ equals a pair of opposite points.

Of course $f^{-1}(z)$ is just a different notation of $S_Z$. So the important detail is: "and $\operatorname{card}(f^{-1}(z))\leq 2$".

To summarize the answer is: Gromov's Waist Inequality almost implies Borsuk-Ulam.

Yashar Mermarian writes here that the answer is yes. And the argument he gives is pretty much the same as the one you already started.

Taking $n=k$ and $\epsilon=\pi/2$ Gromov's waist inequality gives a point $z\in \mathbb{R}^n$ such that $$vol(U_{\pi/2}(S_Z))\geq vol(U_{\pi/2}(S^0))=vol(U_{\pi/2}(\{(1,0,\dots,0),(-1,0,\dots,0)\})=vol(S^n).$$ It follows that $U_{\pi/2}(S_Z)$ has to have the same volume as the sphere. This can only be the case when $S_Z$ contains at least $2$ elements, since the volume of $U_\epsilon(S_Z)$ would be to small otherwise. If it contains $2$ elements and not more, these $2$ elements would have to be antipodal, otherwise the volume of $U_\epsilon(S_Z)$ would be to small again. Although Memarian writes there "is no choice" for $S_Z$ but to "pass through antipodal points", I don't really see how one can rule out that $S_Z$ might consist of more than two points. I hope someone can make this clear.

To summarize this gives the following version of a Bursuk Ulam type result:

If $f:\; S^n\to\mathbb{R}^n$ is continuous, then $f$ maps two antipodal points in $S^n$ to the same point in $\mathbb{R}^n$ or $f$ maps (at least) three points to the same point in $\mathbb{R}^n$.

Yashar Mermarian writes here that the answer is yes. And the argument he gives is pretty much the same as the one you already started.

Taking $n=k$ and $\epsilon=\pi/2$ Gromov's waist inequality gives a point $z\in \mathbb{R}^n$ such that $$vol(U_{\pi/2}(S_Z))\geq vol(U_{\pi/2}(S^0))=vol(U_{\pi/2}(\{(1,0,\dots,0),(-1,0,\dots,0)\})=vol(S^n).$$ It follows that $U_{\pi/2}(S_Z)$ has to have the same volume as the sphere. This can only be the case when $S_Z$ contains at least $2$ elements, since the volume of $U_\epsilon(S_Z)$ would be to small otherwise. If it contains $2$ elements and not more, these $2$ elements would have to be antipodal, otherwise the volume of $U_\epsilon(S_Z)$ would be to small again. Although Memarian writes there "is no choice" for $S_Z$ but to "pass through antipodal points", I don't really see how one can rule out that $S_Z$ might consist of more than two points. I hope someone can make this clear.

To summarize this gives the following version of a Bursuk Ulam type result:

If $f:\; S^n\to\mathbb{R}^n$ is continuous, then $f$ maps two antipodal points in $S^n$ to the same point in $\mathbb{R}^n$ or $f$ maps (at least) three points to the same point in $\mathbb{R}^n$.

$U_{\pi/2}$ for two points which are not antipodal.