# Why these conditions make this map open?

If $A \subset \mathbb{R}^n$ is an open set and $g: A \to \mathbb{R}^n$ is an injective continuously differentiable function such that $\forall x \in A, \, \det g'(x) \neq 0$, does $g(U)$ is open for each $U \subset A$ open? Why? This is about p. 67 of Spivak's Calculus on Manifolds (down of the page), where he says: "the collection of all $g(U)$ is an open cover of $g(A)$".

## 1 Answer

For any $u \in U$, $g'(u)$ is invertible so the inverse function theorem says $g$ is a homeomorphism from a neighbourhood of $u$ to a neighbourhood of $g(u)$. In particular $g(V)$ is open for some open set $V \subseteq U$ containing $u$. Then $g(U)$ is the union of these open sets, so it is open.

• Thank you, Robert Israel!! – João Júnior Jul 15 '12 at 18:14
• Every diffeomorphism is a homeomorphism. Since $g$ is a local diffeomorphism, it is a local homeomorphism and, therefore, sends sufficiently small open subsets of $U$ to open sets. – João Júnior Jul 15 '12 at 18:41