If $B_1\subseteq\mathbb R^d$ and $f$ is a diffeomorphism of $B_1$ onto an open subset of $\mathbb R^d$, then $B_1$ is open

Question

Let

• $d\in\mathbb N$
• $B_1\subseteq\mathbb R^d$
• $\Omega_2\subseteq\mathbb R^d$ be open
• $f:B_1\to\Omega_2$ be a $C^1$-diffeomorphism (in the sense of equation $(2)$ in this question)

Why can we conclude that $B_1$ is open?

This seems to be an application of the inverse function theorem. By definition, $$f=\left.\tilde f\right|_{B_1}\tag2$$ for some $\mathbb R^d$-open neighborhood $\Omega_1$ of $B_1$ and some $\tilde f\in C^1(\Omega_1,\mathbb R^d)$. Now it should hold $$\operatorname{id}_{\mathbb R^d}={\rm D}\left(\tilde f\circ f^{-1}\right)(x_1)={\rm D}\tilde f(x_1)\circ{\rm D}f^{-1}(f(x_1))\tag3$$ for all $x_1\in B_1$. How can we proceed?

Answer 1

$f$ is differentiable so if $g=f^{-1}$, $g$ can be seen as a differentiable map to $\Omega_1$. We have $$\tilde{f} \circ g = \text{id}_{\Omega_2}.$$ Applying the chain rule we see that the jacobi matrix of $g$ is nonsingular at all points of $\Omega_2$. Let $g(y)=x \in B_1$. By the inverse function theorem there exists an open neighbourhood of $x$ such that all points of this neigbourhood are in the image of $g$. But the image of $g$ is contained in $B_1$. Since $x$ was arbitrary this shows $B_1$ can be covered by opens in $\mathbb{R}^d$ and is thus open.