This seems like a very basic result in Algebraic Topology (e.g. it shows up as exercise 0.1.1.1, the very first one, in John Stillwell's "Classical Topology and Combinatorial Group Theory").
I am interested in having a full-blown proof, something that surprisingly I wasn't able to find that easily online to compare to.
Intuitively, its clear that the mapping we're interested in can be found by representing every point using the barycentric co-ordinates in each of the respective n-simplexes and then simply defining the homeomorphism to be the mapping transforming points by maintaining the convex combination of weights (all non-negative, summing to one) but switching out the "basis vectors" corresponding to the vertices of the n-simplex, e.g.:
$$ x = \sum_{i=1}^{n+1} w_i\vec{p_i} \rightarrow f(x) = \sum_{i=1}^{n+1} w_i\vec{q_i}$$
What I'm looking for is to have full chain of reasoning why this is indeed a homeomorphism (e.g. a continuous mapping with a continuous inverse).
One line of reasoning I came up with that seems a bit more convoluted than should be called for and makes use of a generous amount of point-set toplogy goes like this:
- we can represent any point in the "standard" n-simplex in $\mathbb{R}^{n+1}$ by identifying the vertices of the n-simplex with the basis vectors $\vec{e_i}$ and then each point in the n-simplex is simply represented as $\vec{w} = (w_{1}, w_{2}, ..., w_{n+1})$, with the w's corresponding to the non-negative weights summing to 1.
- We can define a mapping from this standard n-simplex in $\mathbb{R}^{n+1}$ to any other n-simplex in $\mathbb{R}^{n}$ (or $\mathbb{R}^{n+1}$ or above..) by $f(\vec{w}) = \sum_{i=1}^{n+1}w_i\vec{p_i}$ (the p vectors representing the vertices of the n-simplex we are mapping into)
- Given that this is a linear transformation (e.g. can be represented as $f(\vec{w}) = P\vec{w}$ where P is the matrix having the $\vec{p_i}$ vectors as columns), it is also continuous, and it is easy to see that it is also injective (keeping in mind we limit the $w_i$'s to be non-negative and sum to one), and so we have a result in topology that tells us this is a homeomorphism (See Corollary here).
- we can now compose the inverse of this mapping (e.g. from one n-simplex P to the standard n-simplex), with the (forward) mapping from the standard n-simplex to the other n-simplex Q, and get the desired homeomorphism (since composition of homeomorphisms results in a homeomorphism).
Is this right? are there much simpler proofs from basic linear algebra/analysis that are usually used here otherwise?