Does every positive ray intersect a deformed simplex? A topological conjecture.

Question

This question is motivated by my partial answer to a different question. I will use $\mathbb R_+$ to denote the set of nonnegative reals.

Consider the standard simplex $\Delta^n=\{(x_0,\dots,x_n)\in\mathbb R_+^{n+1}:x_0+\dots+x_n=1\}.$ We are given a continuous function $f:\Delta^n\to\mathbb R_+^{n+1}$ with the property that it preserves zero coordinates, that is, if $x_i=0$ then $f(x)_i=0.$ Thus vertices of the simplex map to points on the coordinate axes, $1$-faces (edges) map to curves on the coordinate $2$-planes, and so on. I believe the following conjecture is true, but I don't know how to prove it for arbitrary $n$:

Conjecture: For any $y\in\mathbb R_+^{n+1}$, there exists a point $x\in\Delta^n$ such that $f(x)=ay$ for some scalar $a\in\mathbb R_+$. Geometrically, every ray from the origin lying in $\mathbb R_+^{n+1}$ must intersect the surface $S = f(\Delta^n)$.

Here are some examples with $n=1,y=(1,1)$ and $n=2,y=(1,1,1)$ respectively:

These examples suggest that when $y\in\operatorname{int}\mathbb R_+^{n+1}$, the boundary $\partial S=f(\partial\Delta^n)$ "surrounds" the line $\{ay:a\in\mathbb R\}$, so the surface $S$ must intersect the line. Thus I feel the conjecture is essentially topological in nature and should have a natural proof based on something like homotopy theory. Unfortunately, I don't know any homotopy theory.

Answer 1

This can be proved using Brouwer's fixed point theorem. Note that the case that $f(x) = 0$ for some x is trivial, so by the projection $x \mapsto x/\|x\|_1$ we can reduce the problem to whether a mapping of a simplex into itself such that every face is mapped into itself must be surjective. This is indeed the case. Suppose it were not, then there would be some simplex $S$ and a point $x\in S^\circ$ that is not in the image of $f$. There would then be a retraction $r: S\setminus\{x\} \to \partial S$. We can also find a linear mapping $m:S\to S$ that cyclically permutes all the vertices. The composition $m \circ r \circ f$ would then be a continuous mapping of $S$ into itself without fixed points.

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To make this a little more explicit for the present problem, I will assume that $y_i \ne 0$ for all $i$. When this is not the case, the problem can easily be reduced to a lower dimension. For a given $f$, supposing $f(x)$ and $y$ are never dependent, we can then define a function $g:\Delta^n\to\mathbb R_+^{n+1}$ by $$ g(x)_i = f(x)_i / y_i - \min_j f(x)_j / y_j $$ This is clearly a continuous function and $g(x)_i = 0$ when $f(x)_i = 0$. Furthermore, there is always an $i$ for which $f(x)_i/y_i$ is minimal, so $g(x)_i = 0$, but $g$ does not vanish because of the independence of $y$ and $f(x)$. Hence we can define $h:\Delta^n\to\Delta^n$ by $$ h(x)_i = \cases{ g(x)_{n+1}/\|g(x)\|_1 & if $i = 1$ \\ g(x)_{i-1}/\|g(x)\|_1 & otherwise. } $$ Because $g(x)_i = 0$ for some, but not all $i$, there is for every $x$ an $i$ such that $h(x)_i = 0$, but $g(x)_i \ne 0$ and therefore $x_i \ne 0$. It follows that $h(x) \ne x$ for all $x \in \Delta^n$. This contradicts BFPT, so our assumption that $y$ and $f(x)$ are always independent must fail.

Answer 2

Maybe you could use the generalized Jordan curve theorem for $S^n$ to prove this.

Given an embedding $f$ of the $n$-dimensional sphere $S^n$ into $\Bbb R^{n+1}$, then the space $$\Bbb R^{n+1}\setminus f[S^n]$$ obtained by cutting out this embedding from the ambient space will leave exactly two connected components, one bounded and one unbounded.

My idea is that your embedding $f$ of the simplex into $\Bbb R^{n+1}$ can be extended to an embedding $\hat f$ of $S^n$ by mirroring the embedded simplex on all the coordinate (hyper-)planes.

Now assume there is a ray starting in $(0,0)$ going to infinity and not intersecting the simplex $f(\Delta^n)$. This ray also will not intersect the sphere $\hat f(S^n)$ because it only lies in the single positive orthant $\Bbb R^n_+$. Because it starts in $(0,0)$ it belongs to the bounded component of the sphere (proof?). But obviously the ray is undbounded. Contradiction!

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You see that this is just a rough sketch. When I have more time I may formalize it sufficiently. I think the advantage of this informal proof is that it externalizes the heavy homotopy/homology theory to the proof of the generalized Jordan curve theorem.