Embedding of $\mathbb{R}^2 \to \mathbb{R}^3$ with non-parallel tangent planes

Question

I have a qual question here and I'm struggling to get a good starting point. The question asks to construct a smooth proper embedding $f\colon \mathbb{R}^2 \to \mathbb{R}^3$ such that for any distinct $x,y \in \mathbb{R}^2$, the tangent planes to $f(x)$ and $f(y)$ are not parallel.

Here, proper is meant in the sense that the pre-image of any compact set is again compact.

Any hint or starting point would be great. Thanks in advance for your help!

Answer 1

I like $(x,y)\mapsto(x,y,x^2+y^2)$. $\ddot\smile$

Answer 2

To expand on the solution by @Christoph and generate more graph embeddings, you could think about angles of 'pitch' and 'roll' in the sense of airplanes:

In this language, the non-parallelity constraint means that no two tangent planes can have the same pitch and roll. Given a graph of $f: \mathbb{R}^2 \to \mathbb{R}$, the pitch and roll of the plane tangent at $(x_0,y_0,f(x_0,y_0))$ can be equated with $\partial_x f(x_0,y_0)$ and $\partial_y f(x_0,y_0)$, respectively. Thus it suffices to choose $f$ such that $df_{(x_1,y_1)}=df_{(x_2,y_2)}$ if and only if $(x_1,y_1)=(x_2,y_2)$.

Example 1. $f(x,y)= x e^y$ has $\partial_x f(x,y)= e^y$ and $\partial_y f(x,y)= x e^y$. If $d f_{(x_1,y_1)}=df_{(x_2,y_2)}$, then we have $e^{y_1}=e^{y_2}$ and thus $y_1=y_2$. Then $x_1 e^{y_1}=x_2 e^{y_2}=x_2 e^{y_1}$ implies that $x_1=x_2$.

Example 2. If $\partial_x f$ is independent of $y$ and $\partial_y f$ is independent of $x$ and both functions are injective (as functions of $x$ and $y$, respectively), then $df_{(x_1,y_1)}=df_{(x_2,y_2)}$ implies that $\partial_x f(x_1,\cdot)=\partial_x f(x_2,\cdot)$, hence $x_1=x_2$. Similarly, $y_1=y_2$. Note that Christoph's solution falls into this category.