# 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, 2019 at 10:18
• @Max And why should that give a homeomorphism? How do you show that it's injective? Mar 1, 2019 at 10:54
• @stressedout Then you use the inverse function theorem.
– Max
Mar 1, 2019 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? Mar 1, 2019 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, 2019 at 12:39

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.
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 $$p$$.
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.