Can we represent the interior of a polygon as the graph of a continuously differentiable function?

Question

Let $P$ be a polygon in $\mathbb R^3$. Can we find an open set $U\subseteq\mathbb R^2$ and a continuously differentiable function $f:U\to\mathbb R$ such that $$\text{int}(P)=\left\{\left(x,f(x)\right):x\in U\right\}\;?$$ The polygon can be assumed to be "drawable in 2D" and we can assume, that its edges do not cross.

Answer 1

If I understand your question correctly, then yes. A hyperplane in $\mathbb{R}^3$ is the graph of a linear function of two variables on some of the coordinate axes. That is, the projection operator that takes one of the variables to zero will be a diffeomorphism. The catch is that you can't always choose the same projection operator for each hyperplane. For instance, the hyperplane $z=2$ can't be diffeomorphically projected onto anything but the $xy$ plane. (That is, project "along the $z$ axis").

To find the right projection operator, consider a vector $v \in \mathbb{R}^3$ orthogonal to the hyperplane, and project along any axis where $v$ doesn't have zero component in that direction.

Now just find the right projection operator $\pi$, and your polygon $P$ will be a graph of the linear function on $\pi(P) \subset \mathbb{R}^2$. Projections are open maps, so if you start with an interior of a polygon, $\pi(P)$ will be open.

Example: If $(0,0,0)$, $(1,1,0)$, $(1,1,1)$ is the polygon (triangle) given to me, I notice that it lies in the simple hyperplane given by $\{(x,y,z) \mid x - y =0 \}$. In other words, an orthogonal vector to this hyperplane is $(1,-1,0)$. That means I can choose to project along either the $x$ or $y$ axes and obtain a diffeomorphism. If I project along the $x$ axis, I find that my polygon is the graph of the triangle $(0,0)$, $(1,0)$, $(1,1)$ in the $yz$ plane.