I am reading John Lee's Introduction to Topological Manifolds and I am trying to prove Theorem 5.39 (Simplicial Maps are determined by vertex maps. Let $K$ and $L$ be simplicial complexes. Suppose $f_0:K_0 \to L_0$ is any map with the property that whenever $\{v_0,\dots,v_k\}$ are the vertices of a simplex of $K$, $\{f_0(v_0),\dots, f_0(v_k)\}$ are the vertices of a simplex of $L$ (possibly with repetitions). Then there is a unique simplicial map $f: |K| \to |L|$ whose vertex map is $f_0$. It is a simplicial isomorphism if and only if $f_0$ is a bijection satisfying the following additional condition: $\{v_0,\dots, v_k\}$ are the vertices of a simplex of $K$ if and only if $\{f_0,\dots, f_0(v_k)\}$ are the vertices of a simplex of $L$.
An affine map $F: \mathbb{R}^n \to \mathbb{R}^m$ is any map of the form $F(x)=c+A(x)$, where $A$ is a linear map and $c$ is some fixed vector in $\mathbb{R}^m$.
$\sigma=[v_0,\dots, v_k]$ is a $k$-simplex if $\{v_0,\dots, v_k\}$ is an affinely independent set of $k+1$ points in $\mathbb{R}^n$ and $[v_0,\dots v_k]=\{\sum_{i=0}^k t_i v_i : t_i \ge 0\; \text{and} \;\sum_{i=0}^k t_i=1\}.$ Each of the points $v_i$ is called a vertex of the simplex.
Given a set of vertices $\{v_0,\dots, v_k\}$, for which $\{f_0(v_0),\dots f_0(v_k)\}$ are the vertices of a simplex of $L$ (possibly with repetitions), I am trying to use Proposition 5.38 which states that given $\sigma=[v_0,\dots, v_k]$ a $k$-simplex in $\mathbb{R}^n$, and any $k+1$ points $w_0,\dots, w_k \in \mathbb{R}^m$, there is a unique map $f:\sigma \to \mathbb{R}^m$ that is the restriction of an affine map, and takes $v_i$ to $w_i$ for each $i$.
I can prove the theorem if I can show that $f$ takes $\sigma$ onto a simplex of $L$. But I can't show why the affine map maps a simplex onto another simplex. How can we prove this from the definition? I would greatly appreciate some help.
My attempt: The image of $f(\sigma)$ is of the form $\{f(\sum_{i=0}^k t_i v_i): t_i\ge 0, \sum_{i=0}^k t_i =1\}$.
$f(\sum_i t_i v_i) =c+A(\sum_i t_i v_i) = c+ \sum_i t_i A v_i = \sum_i t_i c + \sum_i t_i A v_i = \sum_i t_i(c + Av_i)=\sum_i t_i f(v_i)=\sum_i t_i f_0(v_i),$ because $\sum_i t_i=1$.
So the image of $f(\sigma)$ is $\{\sum_{i=0}^k t_i f_0(v_i): t_i \ge 0 \; \text{and}\; \sum_{i=0}^k t_i=1\}$. But I can't conclude that $f(\sigma)=[f_0(v_0),\dots, f_0(v_k)]$ because the definition requires $\{f_0(v_0),\dots, f_0(v_k)\}$ be affinely independent, and indeed here they may be some repetitions. So assume that $\{f_0(v_0),\dots , f_0(v_l)\}$ are the vertices of a simplex of $L$ by hypothesis of the Theorem. Then by the definition of a vertex, these points are affinely independent, and $[f_0(v_0),\dots, f_0(v_l)]=\{\sum_{i=0}^l t_i f_0(v_i): t_i \ge 0, \sum_{i=0}^l t_i=1\} $ is exactly the set $\{f(\sum_{i=0}^k t_i v_i):t_i\ge 0, \sum_{i=0}^k t_i=1\}$ because in the sum above we can combine the repetitive points together.
Best Answer
A $k$-simplex in $\mathbb R^n$ is the convex hull $\sigma = [v_0,\ldots,v_k]$ of $k+1$ point $v_i \in \mathbb R^n$ which are affinely independent.
Q1. Given an affine map $F : \mathbb R^n \to \mathbb R^m$, is $F(\sigma)$ a simplex in $\mathbb R^m$?
In general no. You have shown in your question that it is the convex hull of the points $F(v_i) \in \mathbb R^m$, thus it is a convex subset of $\mathbb R^m$. But Lee's Proposition 5.38 shows that the $F(v_i)$ can be arbitrary (not necessarily distinct) points of $\mathbb R^m$. For example, consider a $4$-simplex $\sigma = [v_0,\ldots,v_4] \subset \mathbb R^n$ for $n \ge 4$. Then there is an affine map $F : \mathbb R^n \to \mathbb R^2$ such that $F(v_i) = w_i$ with
$$w_0 = (0,0), w_1 = (1,0), w_2 = (1,1), w_3 = (0,1), w_4 = (1/2,1/2) .$$
$F(\sigma)$ is the square with vertices $w_0, w_1, w_2, w_3$. The point $w_4$ lies in the interior of of the square.
Q2. If the $v_i \in \mathbb R^n$ are affinely independent, does there exist a unique affine map $F : \mathbb R^n \to \mathbb R^m$ which maps the $v_i$ to the prescribed $w_i$?
This is is true only when $n = k$. As in Lee's proof it suffices to consider the case $v_0 = w_0 = 0$. If $n = k$, then $\{v_1,\ldots,v_k\}$ is a basis of $\mathbb R^k$ and we get uniqueness. If $n > k$, we can extend the linearly independent set $\{v_1,\ldots,v_k\}$ to a basis $B$ of $\mathbb R^k$. Then each $b \in B \setminus \{v_1,\ldots,v_k\}$ can be mapped arbitrarily into $\mathbb R^m$ which produces infinitely many $F$ with $F(v_i) = w_i$.
Q3. So what does Proposition 5.39 say?
The essence is this:
By Proposition 5.38 there exists a unique $f: \sigma \to \mathbb R^m$ which is restriction of an affine $F : \mathbb R^n \to \mathbb R^m$ such that $F(v_i) = f_0(v_i)$.
If the set $f_0(\{v_0,\ldots,v_k\}$, which can be written in the form $\{w_0,\ldots,w_r\}$ with pairwise distinct $w_i$, is the vertex set of an $r$-simplex in $\mathbb R^m$ (which means that the $w_i$ are affinely independent), then $f(\sigma)$ is the simplex $[w_0,\ldots,w_r]$.
In fact, we know that $f(\sigma)$ is the convex hull of the set $\{w_0,\ldots,w_r\}$ .