Linear map of extrem points and its injectivity, surjectivity

Question

Before entering my problem, let me review some related results:

Suppose $\mathcal{S}$ is a convex hull of finite points: $\mathcal{S}= \operatorname{conv}(x_1,x_2,\ldots,x_m)$, then

1. By "Convex hull of extreme points" , we know $\mathcal{S}$ is the convex hull of its extreme points, which is the subset of $\{x_1,\ldots, x_m\}$.
2. By "preservation of extreme points under linear transformation", if the transformation is linear, let $A$ = {the extreme points of the image of a compact convex set}, then the preimage of $A$ must be the extreme points of that compact convex set. But the inverse in not true. The inverse is true only under such transformation is injective and surjective.

My question is:

Suppose

1. We have two sets $A,B$ with $L: A\rightarrow B$, which is linear.
2. $A,B$ are both convex, compact and both sets have at least one extreme point.
3. Suppose $L$ maps extreme points in $A$ one-to-one and onto to extreme points in $B$.

So obviously, $A$ and $B$ are convex hull of their extreme points.

Question: Is $L$ a one-to-one and onto linear map for $A\rightarrow B$?

If not, could you please provide an counter-example?

Answer 1

I think the assertion is false.

Let $A \subset \mathbb{R}^3$ be a tetrahedron and let $L$ be a projection onto a two-dimensional subspace, such that $B = L\,A$ is a quadrangle.