Homeomorphism of $\Delta^n$ into itself that switches interior points

Question

Let $\Delta^n$ be the standard $n$-symplex. Let $x, y$ two interior points of $\Delta^n$. How can I prove that there exists an homeomorphism of $\Delta^n$ into itself that maps $x$ to $y$?

I can see that this map exists, but I can't write it down.

Answer 1

This is a general fact for manifolds. (There is a homeomorphism taking one point to another, if they are in the same path connected component.)

One way to do it is as follows: Build a little tube around a path connecting the two points, and do so that the tube is topologically a ball. Then you just need to prove that there is a homeomorphism of a ball fixing the boundary but sending one interior point $x$ to another given interior point $y$.

Edit: After some thought, I think I have a nice way to do this:

You identify the interior of the ball with $R^n$. Then the additive group of $R^n$ by translations is obviously transitive on points, but since it fixes classes of directed parallel lines, the action on the sphere at infinity is trivial. This should give a homeomorphism of a ball doing what you want.

(Earlier argument: One way to write this down is by ensuring that they both lie on the sphere of radius r, then extending a rotation moving one to the other two a homeomorphism of the whole disc by slowly rotating by less as you go inwards and outwards.)

Of course in your case the simplex is already topologically a ball, so you can skip the first step.

Answer 2

Let $\Delta$ have vertices $v_0,…,v_n$. The simplex $\Delta$ is the convex hull of $v_0,…,v_n$, denoted $\Delta = [v_0,…,v_n]$. This set consists of all convex combinations $t_0 v_0+…+t_n v_n$ where the $t_i$'s are non-negative and sum to $1$. These numbers $t_0,…,t_n$ are called the "barycentric coordinates" of the point.

The co-dimension one faces of $\Delta$ are all simplices of one dimension lower, one face for each $i=0,..,n$, namely $\Delta_i =$ the convex hull of $\{v_0,…,v_n\}-\{v_i\}$.

Consider $p,q \in \text{interior}(\Delta)$. Let $\Delta_{i,p}$ denote the convex hull of the vertices of $\Delta_i$ together with the point $p$, and similarly for $\Delta_{i,q}$.

Define $f : \Delta \to \Delta$ as follows: for each $i=0,..,n$ the map $f$ takes $\Delta_{i,p}$ to $\Delta_{i,q}$ preserving barycentric coordinates. One can check that $f$ is well-defined and continuous and takes $p$ to $q$: the key observation is that for each $x \in \Delta_{i,p} \cap \Delta_{j,p}$ the nonzero barycentric coordinates of $x$ in those two simplices are precisely the same coordinates with the same values.

One can also reverse the process, constructing the inverse of $f$ by taking $\Delta_{i,q}$ to $\Delta_{i,p}$ preserving barycentric coordinates. Thus $f$ is a homeomorphism.