---
tag: 9WR72RV403
kind: proof
name: Proof that a C^1 map with invertible derivative is a local homeomorphism and an open map
msc: ["26B10"]
proves: 0Z8CSYVHDJ
uses: [0K9HSJFCMJ, 0135XZZQ8M, 0KXSK8TEEB, 0S9EK5MQPE, 05GZ11PWRY, 0SQ2PC596Y]
sources:
  - url: https://en.wikipedia.org/wiki/Inverse_function_theorem
    locator: Several variables
---

## Proof

We apply the inverse function theorem at each point.

1. Let $U$ and $f$ be as in the statement, and let $a \in U$. Since $Df(a)$ is invertible, its Jacobian determinant is not zero.
2. By the [[inverse function theorem|0K9HSJFCMJ]] with $k = 1$, there are open sets $V \subseteq U$ and $W \subseteq \R^n$ with $a \in V$ and $f(a) \in W$. The map $f$ restricted to $V$ is a bijection onto $W$. Its inverse $g \colon W \to V$ is of class $C^1$. So $f(V) = W$ is open.
3. A function that is [[differentiable|0135XZZQ8M]] at a point $p$ is [[continuous|0S9EK5MQPE]] at $p$. Indeed, $\norm{F(p + h) - F(p)} \le \norm{F(p + h) - F(p) - L h} + \norm{L} \norm{h}$ with $L = DF(p)$, and both terms tend to $0$ as $h \to 0$.
4. The maps $f$ and $g$ are of class $C^1$, so they are [[differentiable|0KXSK8TEEB]] at each point of their open domains. By step 3, $f$ restricted to $V$ and $g$ are continuous.
5. So $f$ restricted to $V$ is a continuous bijection onto $f(V)$ with continuous inverse. That is, it is a [[homeomorphism|05GZ11PWRY]] onto the open set $f(V)$.
6. Now let $O \subseteq U$ be open, and let $b \in f(O)$, say $b = f(a)$ with $a \in O$.
7. The restriction of $f$ to the open set $O$ is of class $C^1$ with invertible derivative at each point. By steps 1 to 5 applied to it, there is an open $V_a \subseteq O$ with $a \in V_a$ and $f(V_a)$ open.
8. Then $b \in f(V_a) \subseteq f(O)$. So each point of $f(O)$ has an open neighbourhood inside $f(O)$, and $f(O)$ is open.
9. Hence $f$ is an [[open map|0SQ2PC596Y]].

## Idea

The inverse function theorem gives, around each point, an open set on which $f$ has a $C^1$ inverse. Differentiable maps are continuous, so this local inverse is continuous and $f$ is a homeomorphism there. An open set is a union of such neighbourhoods, so its image is a union of open sets.