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Why is the graph, $G$, of a smooth function $f:\mathbb R^2 \to \mathbb R^3$ necessarily an "embedded surface"?

Thank you.

Comment: I am starting to thing that there is a typo in the text. I think $f:\mathbb R^2 \to \mathbb R$ so that the graph $G=(x,y,f(x,y))$. Assuming this is true, does the question make sense now?

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    @MarianoSuárez-Alvarez: Sorry, deleted. – 2012-03-11

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Yes, you want the graph in $\mathbb{R}^3$ of a function $f:\mathbb{R}^2\to\mathbb{R}$. The easy way to see that this is an embedded surface is to define the map $F:\mathbb{R}^2\to\mathbb{R}^3$ via $F(x,y) = (x,y,f(x,y))$ and check that $F$ is an embedding: it is

  • injective,
  • smooth, and
  • $dF$ is everywhere full rank.

The first two are easy to check. The third follows from the tact that $dF = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \\ \end{pmatrix}$ always has trivial kernel.

(In fact, the graph of a smooth function is actually a two-dimensional submanifold of $\mathbb{R}^3$, which is slightly stronger than noting that it is an embedded surface. Observe that we can produce new coordinates on $\mathbb{R}^3$ by "flattening" the graph down to the $xy$-plane. The change-of-coordinate map is $\phi(x,y,z) = (x,y,z - f(x,y))$; it's a diffeomorphism, which means that the graph of $f$ has a coordinate chart in which it is the $xy$-plane, i.e., it is a two-dimensional submanifold of $\mathbb{R}^3$.)