# Prove there is an open set $\widetilde W\subset W$ containing $q$ such that $\sigma(\widetilde W)$ is a smooth surface

Let $$W\subset\mathbb{R}^2$$ be open $$\sigma:W\to \mathbb{R}^3$$ be a smooth map. Let $$\sigma_u(q)\times\sigma_v(q)\not=0$$ for some $$q\in W$$. Prove there is an open set $$\widetilde W\subset W$$ containing $$q$$ such that $$\sigma(\widetilde W)$$ is a smooth surface.

I believe this should be application of inverse function theorem, since essentially you should have that there is an open neighborhood of $$p,\widetilde W$$ where $$\sigma_u\times\sigma_v\not=0$$, $$\sigma$$ is smooth and regular in this neighborhood and there exists an open neighborhood $$V\subset\mathbb{R}^3$$, such that every point $$q\in V$$, $$q\in\sigma(\widetilde W)$$. So $$V$$ is a smooth surface.

My issue is I'm having some trouble seeing where I use the theorem. I need to define a function to actually use it on and I'm not sure how I should do that.

Below is my definition of a smooth surface:

If $$S$$ is a surface, an allowable surface patch for $$S$$ is a regular surface patch $$\sigma:U\to \mathbb{R}^3$$ such that $$\sigma$$ is a homeomorphism from $$U$$ to an open subset of $$S$$.

A smooth surface is a surface $$S$$ such that, for any point $$p\in S$$, there is an allowable surface patch $$\sigma$$ such that $$p\in \sigma(U)$$.

• Try composing $\sigma$ with a projection to one of the coordinate planes (say, pick coordinate plane for which the "remaining" component of $\sigma_u\times\sigma_v$ is non-zero). – Max Mar 1 at 10:18
• @Max And why should that give a homeomorphism? How do you show that it's injective? – stressed out Mar 1 at 10:54
• @stressedout Then you use the inverse function theorem. – Max Mar 1 at 10:59
• @Max Yup, but provided that what you say is true. I can't immediately see why it's true. It's not intuitive. Have you done the calculation to see that the Jacobian of your proposed function is invertible? – stressed out Mar 1 at 11:01
• @stressedout sure. Projection is linear, so its Jacobian is itself i.e. projection. Thus we are talking about taking two rows of Jacobian of $\sigma$ and claiming resulting 2 by 2 matrix is full rank; but the determinant of that matrix is precisely the "complementary" component of $\sigma_u\times \sigma_v$. Intuitively, the component of normal vector being non-zero says the tangent plane is non-vertical, and so projects to the base in 1-1 way. – Max Mar 1 at 12:39

## 1 Answer

You can't use the inverse function theorem directly because the inverse function theorem is applicable only when the dimensions of the domain and the co-domain of the function which we want to apply it to are the same. Since $$W \subset \mathbb{R}^2$$ is two-dimensional while your co-domain is three-dimensional, the inverse function theorem cannot be applied directly.

However, you can use the following statement which can be proved using the constant rank theorem. The constant rank theorem itself is proven using the inverse function theorem

Lemma: Let $$f: U\subseteq \mathbb{R}^n \to \mathbb{R}^m$$ be a $$C^1$$ function on an open set $$U$$ where $$n \leqslant m$$ such that $$\mathrm{rank}{Df}=n$$ at some $$p\in U$$. Then $$f$$ is injective in a neighborhood of $$p$$.

For the proof, see my question here. I have given a proof using the constant rank theorem and hopefully, someone can come up with a more elementary proof. In any case, the proof of the constant rank theorem using the inverse function theorem is standard and it's found in almost every standard textbook in analysis like Rudin's Mathematical Analysis or Pugh's Real Mathematical Analysis, etc.

Anyway, the assumption that $$\sigma_u(q)\times\sigma_v(q)\not=0$$ tells us that the two vectors $$\sigma_u(q)$$ and $$\sigma_v(q)$$ are linearly independent, i.e. the Jacobian matrix of $$\sigma$$ at $$q$$, $$D_q(\sigma)$$, has rank $$2$$. Using the lemma I've highlighted, you will conclude that $$\sigma$$ is locally injective near $$q$$, i.e. there exists $$\widetilde{W} \subseteq W \subset \mathbb{R}^2$$ such that $$\sigma: \widetilde{W} \to \sigma(\widetilde{W}) \subseteq \mathbb{R}^3$$ is a homeomorphism (why?). So, $$\sigma(\widetilde{W})$$ is a smooth surface as you wanted to demonstrate.