On page 32 of Ziegler's lectures, he wants to show that every $\mathcal{V}$-polyhedron is an $\mathcal{H}$-polyhedron. Ziegler defines the $\mathcal{V}$-polyhedron as the Minkowski sum of a convex hull of a finite set $V$ and the cone generated by another finite set $Y$:
$$\mbox{conv}(V)+ \text{cone}(Y) = \left\{ (\lambda_1 v_1 + \dots + \lambda_n v_n) + (c_1y_1 + \dots + c_m y_m): \sum_{i=1}^n \lambda_i = 1 \ \text{and} \ \lambda_i, c_i \geq 0 \right\}$$
He defines the $\mathcal{H}$-polyhedron as the intersection of some closed half-spaces. To demonstrates that every $\mathcal{V}$-polyhedron is an $\mathcal{H}$-polyhedron, he notes that every $\mathcal{V}$-polyhedron,
$$\mbox{conv}(V) + \mbox{cone}(Y) = \left\{ \textbf{x} \in \mathbb{R}^d: \exists \textbf{t} \in \mathbb{R}^n, \textbf{u} \in \mathbb{R}^{n'}: \textbf{x} = V\textbf{t} + Y\textbf{u}, \textbf{t} \geq 0, \textbf{u} \geq 0, \ \text{and} \ \mathbb{1}\textbf{t} = 1 \right\}$$
can be interpreted as the projection of a set
$$ \left\{ (\textbf{x}, \textbf{t}, \textbf{u}) \in \mathbb{R}^{d+n+n'}: \textbf{x} = V \textbf{t} + Y \textbf{u}, \textbf{t} \geq 0, \textbf{u} \geq 0, \ \text{and} \ \mathbb{1}\textbf{t} = 1 \right\}$$
that is quite clearly an $\mathcal{H}$-polyhedron. I have two questions.
What is the projection of a polytope? Is it simply a linear map $f: \mathbb{R}^n \to \mathbb{R}^n$ such that $f \circ f = f?$
Why can the first set be interpreted as a projection of the second set?
I don't quite see the reasoning here and I would totally appreciate any help. Thanks in advance.