Suppose we are given a simplex $X=\{ (x_1,x_2)\in R^2: x_1,x_2\geq 0, x_1+x_2\leq 1\}$ with the extreme points $(1,0)$,$(0,1)$ and $(0,0)$
Consider the product set of $2$-copies of $X$, $X^1\times X^2$. We define a "sum" operation $ T:(x^1,x^2)\to y\in R^4$ by: for every $(x^1,x^2)\in X^1\times X^2$,
$y_{11}=x_1^1+x_1^2$
$y_{12}=x_1^1+x_2^2$
$y_{21}=x_2^1+x_1^2$
$y_{22}=x_2^1+x_2^2$
Denote $Y$ the set of all $y$ given this operation.
If we go over all pairs of extreme points in $X_1$ and $X_2$, we obtain
$Y^*=\{(2,1,1,0), (1,2,0,1),(1,1,0,0),(1,0,2,1),(0,1,1,2),(0,0,1,1),(1,0,1,0),(0,1,0,1),(0,0,0,0 )\}$
It seems to me that $ Y^*$ always contains all extreme points of $Y$ (since $Y$ is obtained by a projection of $X^1\times X^2$). Moreover, it can be verified that every point in $Y^*$ is an extreme point of $Y$, which seems unusual as $T$ is not invertible.
My question is that why this would be the case and could this result generalize to $K$-copies $X^1\times...\times X^{K} $?
Thanks.