Let $g: S^n \to R^n$ be a continuous odd function (i.e. $g(-x)=-g(x)$ for all $x$). [The Borsuk-Ulam theorem](https://en.wikipedia.org/wiki/Borsuk%E2%80%93Ulam_theorem) implies that $g$ has a zero, i.e. there is an $x$ such that $g(x)=(0,0,...,0)$.


Suppose $g$ is (1,1,...,1) on the positive orthant (i.e. when all its $n$ arguments are non-negative) and (-1,-1,...,-1) on the negative orthant. Is this true that for every constant $r\in [-1,1]$, there is an $x$ such that $g(x)=(r,r,...,r)$?


For $n=1$, this is obviously true by the intermediate value theorem. Under what conditions is it true for $n>1$?


The question seems closely related to the [Poincare-Miranda theorem](https://en.wikipedia.org/wiki/Poincar%C3%A9-Miranda_theorem#References), which is a generalization of the IVT to multi-dimensional cubes, but so far I haven't found the connection.



Generalization of Borsuk-Ulam to arbitrary ratio
I believe I have a valid counterexample for $n > 1$, unfortunately I don't have the expertise to be certain. I would glad if someone would expand or refute the following.


Take $n=2$ to start. Now we are going to create a pair of functions $f\_1(x),f\_2(x)$ with $g(x)=(f\_1,f\_2)$ which don't satisfy the property. First take $f=f\_1'=f\_2'$ to be a simple **bipolar** function such that $f$ is $0$ on a great circle and has a maximum at $x\_m = (1,1)$ and minimum at $(-1,-1)$ s.t. $g(x\_m) = M < 1$. 


Now we modify $f\_1'$ into $f\_1$ to have a continuous bump around $x\_m$ of radius $r\_1$ and value $1$, that is, $g(x) \approx (f\_1',\*)$ when $|x\_m \cdot x| \le 1-h(r\_1)$ and $g(x) \approx (1,\*)$ when $|x\_m \cdot x| \gt 1-h(r\_1)$, and accordingly we modify $f\_2$ to have a bump around $x\_m$ of radius $r\_2$ and value $1$, that is, $g(x) \approx (\*,f\_2')$ when $|x\_m \cdot x| \le 1-h(r\_2)$ and $g(x) \approx (\*,1)$ when $|x\_m \cdot x| \gt 1-h(r\_2)$. So that we have $g=(f\_1,f\_2)$.


Now if $h(r\_1) \ll h(r\_2) \ll 1$, then $f\_1$ assumes almost all the values in $[M,1]$ near a circle of radius $r\_1$, while $f\_2$ assumes almost all values in $[M,1]$ near a circle of radius $r\_2$. Clearly, those concentric circles don't intercept. We conclude there exists $y \in [M,1]$ such that $g(x)=(y,y)$ has no solution.


I could generalize this, if I'm not missing something.


