Prove that the image of a regular parametrization is open in the surface

Question

Given a regular parametrization $\mathbf{x}:U \rightarrow \mathbf{x}(U)\subseteq \Sigma$, where $U \subseteq \mathbb{R}^2$ is an open set and $\Sigma \subseteq \mathbb{R}^3$ is a surface, I want to prove that $\mathbf{x}(U)$ is open with respect to the subspace topology of $\Sigma$, using the fact that $\mathbf{x}:U \rightarrow \mathbf{x}(U)$ is a homeomorphism.

I tried to follow mainly two ways:

1. Use the characterization of open sets in subspace topology ($A \subseteq \Sigma$ is open in $\Sigma \Leftrightarrow \exists A' \subseteq \mathbb{R}^3$ open set such that $A=A' \cap \Sigma)$
2. Use the characterization of open sets as neighborhoods of their points

How can I proceed?

Answer 1

Given a regular parametrization $\mathbf{x}: U \rightarrow \mathbf{x}(U) \subseteq \Sigma$ where $U \subseteq \mathbb{R}^2$ is an open set and $\Sigma \subseteq \mathbb{R}^3$ is a surface, we want to prove that $\mathbf{x}(U)$ is open with respect to the subspace topology of $\Sigma$.

1. Regularity of $\mathbf{x}$: The regularity condition implies that the differential $D\mathbf{x}(u)$ is injective for all $u \in U$. In other words, the Jacobian matrix $\frac{\partial \mathbf{x}}{\partial u}(u)$ has full rank for every $u \in U$. We can write this as: $$ \text{rank}\left(\frac{\partial \mathbf{x}}{\partial u}(u)\right) = 2 \quad \text{for all } u \in U $$

2. Inverse Function Theorem (IFT): The IFT states that if $F: U \to \mathbb{R}^2$ is a $C^1$ mapping with continuous partial derivatives and $u_0 \in U$ such that $\text{rank}\left(\frac{\partial F}{\partial u}(u_0)\right) = 2$, then there exist open neighborhoods $V$ of $u_0$ in $U$ and $W$ of $F(u_0)$ in $\mathbb{R}^2$ such that $F: V \to W$ is a diffeomorphism.

3. Proof of openness using the IFT: Let $\mathbf{p} \in \mathbf{x}(U)$. This means that there exists $u_0 \in U$ such that $\mathbf{x}(u_0) = \mathbf{p}$. We want to show that there exists an open neighborhood $N$ of $\mathbf{p}$ in $\Sigma$ such that $N \subseteq \mathbf{x}(U)$.

Since $\mathbf{x}$ is regular, we have $\text{rank}\left(\frac{\partial \mathbf{x}}{\partial u}(u_0)\right) = 2$. By the IFT, there exist open neighborhoods $V$ of $u_0$ in $U$ and $W$ of $\mathbf{x}(u_0)$ in $\Sigma$ such that $\mathbf{x}: V \to W$ is a diffeomorphism.

Let $N = \mathbf{x}(V)$. We claim that $N$ is an open neighborhood of $\mathbf{p}$ in $\Sigma$. To prove this, we need to show that for every $\mathbf{q} \in N$, there exists an open ball centered at $\mathbf{q}$ that is contained in $N$.

Let $\mathbf{q} \in N$. Since $\mathbf{x}: V \to W$ is a diffeomorphism, there exists a unique $u \in V$ such that $\mathbf{x}(u) = \mathbf{q}$. Since $V$ is open, there exists an open ball $B$ centered at $u$ that is contained entirely in $V$. The image of this open ball under $\mathbf{x}$, $\mathbf{x}(B)$, is an open set in $\mathbf{x}(U)$ since $\mathbf{x}$ is a homeomorphism onto its image. Moreover, $\mathbf{q} = \mathbf{x}(u)$ is an element of $\mathbf{x}(B)$.

Therefore, for every $\mathbf{q} \in N$, there exists an open ball centered at $\mathbf{q}$ that is contained entirely in $\mathbf{x}(U)$. Thus, $N$ is an open neighborhood of $\mathbf{p}$ in $\Sigma$.

Since $\mathbf{p}$ was an arbitrary point in $\mathbf{x}(U)$, we have shown that every point in $\mathbf{x}(U)$ has an open neighborhood in $\Sigma$ contained entirely in $\mathbf{x}(U)$. Hence, $\mathbf{x}(U)$ is open with respect to the subspace topology of $\Sigma$. $\square$