# Applying Inverse Function Theorem to open set $A$ to prove $f^{-1}:f(A) \to A$ is a differentiable map

This is the first question in Spivak's Calculus on Manifolds from the chapter on the Inverse Function Theorem.

Problem 2-36. * Let $A \subset \mathbb{R}^n$ be an open set and $f : A \to \mathbb{R}^n$ a continuously differentiable $1$-$1$ function such that $\det f'(x) \neq 0$ for all $x$. Show that $f(A)$ is an open set and $f^{-1} : f(A) \to A$ is differentiable. Show also that $f(B)$ is open for any open set $B \subset A$.

To prove that $f^{-1}:f(A) \to A$ is differentiable, I began by applying the Inverse Function Theorem to some $x\in A$, which gives an inverse function from some subset of $f(A)$, lets call it $W_{f(x)}$ to a subset of $A$, called $V_x$, thus giving $f^{-1}:W_{f(x)} \to V_x$ as a differentiable map. How then is $f^{-1}:f(A) \to A$ differentiable? Do I generate inverse functions for each $y\in f(A)$ and define $f^{-1}:f(A) \to A$ as a piecewise function such that $f^{-1}:W_{f(x_1)} \to V_{x_1}$ is the function that applies when we are mapping an element that's contained in $W_{f(x_1)}$? Or, do I just need one of the differentiable maps $f^{-1}:W_{f(x)} \to V_x$ to prove this?

Thanks in advance.

• If $f$ is locally invertible, that means it is locally an open map. Use that $f(\cup A_\alpha) = \cup f(A_\alpha).$ – Will M. Jun 12 '18 at 22:52
• $f^{-1} : f(A) \to A$ is differentiable at every $f(x) \in f(A)$ since we have neighbourhood $W_{f(x)}$ such that $f^{-1}|_{W_{f(x)}} : W_{f(x)} \to V_{x}$ is differentiable at every point in $W_{f(x)}$. – kelvinn aja Jun 13 '18 at 1:43
• @Sou Thanks! that clears it up – john fowles Jun 13 '18 at 14:26

## 1 Answer

From the comments above.

The function $f^{-1} : f(A) \to A$ is said to be differentiable if it is differentiable at each point in the domain, that is, at each $f(x) \in f(A)$. Since for each point we have a neighbourhood $W_{f(x)} \subseteq f(A)$ of $f(x)$ and a neighbourhood $V_x \subseteq A$ of $x$ such that $$f^{-1}|_{W_{f(x)}} : W_{f(x)} \to V_x$$ is differentiable, we have shown that $f^{-1} : f(A) \to A$ is differentiable.