Image of an extreme point by an affine map

Question

Let $C$ be a convex subset of an affine space $E$.

Is it true that the image of an extreme point of $C$ by an affine map $f: E \longrightarrow F $ is an extreme point of $f(C)$ ?

Answer 1

The projection function

$$ f\;:\;\mathbb R^2\to\mathbb R\qquad\;\; f(x,y):=x $$

is an affine function (even linear) which transform the unit closed disk $B\big((0,0);1\big)$ into the interval $[-1,1]$, both convex sets. The point $(0,1)$ is an extreme point of $B$, but $f(0,1)=0$ is not an extreme point of $[-1,1]$.

But:$\;$ If $f$ is bijective, i.e. it is an affinity (and therefore $E$ and $F$ have the same dimension), extreme points of $C$ go into extreme points of $f(C)$. Indeed, if $P$ is an extreme point of $C$, let suppose that $f(P)$ is not an extreme point of $f(C)$. Then there are distinct points $Q_1,Q_2\in f(C)$ and $\lambda\in\mathbb R$ with $0<\lambda<1$, such that (in barycentric coordinates)

$$ f(P)=\lambda Q_1+(1-\lambda)Q_2\,. $$

Hence

$$ P=f^{-1}(f(P))=\lambda f^{-1}(Q_1)+(1-\lambda)f^{-1}(Q_2), $$

and $P$ would not be an extreme point of $C$, being $f^{-1}(Q_1)$ and $f^{-1}(Q_2)$ distinct points of $C$.