Invariance of domain for smooth functions Let $f \colon U \to \mathbb R^n$ ($U \subset \mathbb R^n$ open) be of class $C^1$ and injective. Apparently there is an easy proof to show that $f(U)$ is open. In general it follows from the Invariance of domain theorem.

Does someone know that proof?
Proof of a somewhat weaker version:

Say $0\in U$ and $f(0)=0$. We need to show that if $||y||$ is small enough then $y=f(x)$.

Since $f$ is injective, the derivative $A=f'(0)$ is nonsingular; now considering $A^{-1}\circ f$ in place of $f$, we may assume $f'(0)=I$.

Edit: Damn. The example $n=1$, $f(t)=t^3$ shows that $f$ injective does not imply that $f'$ is nonsingular. Complex analysis makes things seem too easy. So assume that $f'(x)$ is nonsingular...

Since $f\in C^1$ there exists $r&gt;0$ so that $$||f'(x)-I||&lt;1/2\quad(||x||&lt;r).$$I'm going to be sloppy and assume that $||f'-I||&lt;1/2$ throughout the construction; after it's over  you can go  back and see how small $||y||$ needs to be to ensure everything stays inside this ball.

Let $F(x)=f(x)-y$. We look for a zero of $F$ using a crude version of Newton's method.

Note first that if $||x||,||x'||&lt;r$ then $$||(F(x)-F(x'))-(x-x')||&lt;1/2||x-x'||,\quad(*)$$because $(F(x)-F(x'))-(x-x')$ is the integral of $F'-I=f'-I$ over the segment from  $x'$ to $x$.

Let $x_0=0$ and $$x_{n+1}=x_n-F(x_n).$$Now $$F(x_{n+1})=(F(x_{n+1}-F(x_n))-(x_{n+1}-x_n),$$so (*) shows that $$||F(x_{n+1})||&lt;1/2||x_{n+1}-x_n||=1/2||F(x_n)||.$$So $F(x_n)\to0$. And also $||x_{n+1}-x_n||\to0$ geometrically, so $x_n\to x$.

Cleaning up: Inequalities above show that $||x_n||&lt;C||x_1-x_0||=C||y||$ where $C=2$ or $4$ or something, so if $||y||&lt;r/C$ then $||x_n||&lt;r$ for alll $n$, hence the inequalities actually hold.
Proof of a somewhat weaker version:

Say $0\in U$ and $f(0)=0$. We need to show that if $||y||$ is small enough then $y=f(x)$.

Since $f$ is injective, the derivative $A=f'(0)$ is nonsingular; now considering $A^{-1}\circ f$ in place of $f$, we may assume $f'(0)=I$.

Edit: Damn. The example $n=1$, $f(t)=t^3$ shows that $f$ injective does not imply that $f'$ is nonsingular. Complex analysis makes things seem too easy. So assume that $f'(x)$ is nonsingular...

Since $f\in C^1$ there exists $r&gt;0$ so that $$||f'(x)-I||&lt;1/2\quad(||x||&lt;r).$$I'm going to be sloppy and assume that $||f'-I||&lt;1/2$ throughout the construction; after it's over  you can go  back and see how small $||y||$ needs to be to ensure everything stays inside this ball.

Let $F(x)=f(x)-y$. We look for a zero of $F$ using a crude version of Newton's method.

Note first that if $||x||,||x'||&lt;r$ then $$||(F(x)-F(x'))-(x-x')||&lt;1/2||x-x'||,\quad(*)$$because $(F(x)-F(x'))-(x-x')$ is the integral of $F'-I=f'-I$ over the segment from  $x'$ to $x$.

Let $x_0=0$ and $$x_{n+1}=x_n-F(x_n).$$Now $$F(x_{n+1})=(F(x_{n+1}-F(x_n))-(x_{n+1}-x_n),$$so (*) shows that $$||F(x_{n+1})||&lt;1/2||x_{n+1}-x_n||=1/2||F(x_n)||.$$So $F(x_n)\to0$. And also $||x_{n+1}-x_n||\to0$ geometrically, so $x_n\to x$.

Cleaning up: Inequalities above show that $||x_n||&lt;C||x_1-x_0||=C||y||$ where $C=2$ or $4$ or something, so if $||y||&lt;r/C$ then $||x_n||&lt;r$ for alll $n$, hence the inequalities actually hold.