Suppose $A$ and $B$ are both subsets of $\mathbb{R}^n$ and convex. Let \begin{equation*} C=\left\{z\in \mathbb{R}^n \vert z=\alpha x+\left(1-\alpha\right)y, x \in A, y\in B, \alpha \in \left[0,1\right]\right\}, \end{equation*} that is just the convex hull of $A \cup B$.
Suppose there is a one-to-one mapping between $A$ and $B$, denoted by $f$. Consider the following set where the convex combination is more restrictive: \begin{equation*} D=\left\{z\in \mathbb{R}^n \vert z=\alpha x+\left(1-\alpha\right)y, x \in A, y\in B, y=f\left(x\right), \alpha \in \left[0,1\right]\right\}. \end{equation*}
Apparently $D\subseteq C$. I was wondering whether $D=C$. I tried to construct a counterexample by drawing graphs but failed.
A counterexample in two dimensions: Two parallel segments $$ A = \{ (x, 0) \mid 0 \le x \le 1 \} \\ B = \{ (x, 1) \mid 0 \le x \le 1 \} $$ and the function $f:A \to B$, $f(x, y) = (1-x, 1)$.
The set $D$ consists of all segments from points $(x, 0) \in A$ to $(1-x, 1) \in B$, and does not contain the point $(0, 1/2)$, which is in the convex hull of $A \cup B$.