Consider a (convex, compact, oriented) k-polyhedron $P\subset \mathbb{R}^N$, and $U\subset \mathbb{R}^N$ any $n$-dimensional linear subspace. Let $\pi_U:\mathbb{R}^n\to U$ denote the orthogonal projection onto $U$. Is it true that $\pi_U(P)$ is again a polyhedron? If so, then given any $l$-cell $e\in\pi_U(P)$, is there any algorithm or condition to tell how many $l$-cells of $P$ are in $\pi_U^{-1}(e)$?
To provide context (and to ask a third question), my question comes from the claim that $\pi_U$ induces a map ${\pi_U}_\#:\mathscr{P}_k(\mathbb{R}^n)\to \mathscr{P}_k(U)$. Here, given any euclidean space $E$, we define the (integer) $k$-polyhedral chain complex as $$\mathscr{P}_k(E):=\mathbb{Z}[\text{k-polyhedra}]/\sim,$$ where $P\sim -\bar{P}$ (the bar denotes reversed orientation), and $S\sim P+Q$ if $P,Q$ glue to $S$ along their boundaries (respecting orientation).
Now, I'm a bit confused on the image of ${\pi_U}_\#$. For example, take $N=3$, $U$ to be the $xy$-plane, and $P=2-\text{skeleton of }[0,1]^3=\text{faces of }[0,1]^3$. Let $\sigma$ denote the face of $P$ lying on the $xy$-plane. Under the projection $\pi_U$, the image of the vertical (wrt U) faces of $P$ are edges of $\sigma$, while $\sigma$ and the face parallel to $\sigma$ project to $\sigma$. Therefore, in this example, is ${\pi_U}_\#(P)=\sigma$, or ${\pi_U}_\#(P)=2\sigma$?