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Let $f\in \mathcal C^1(\mathbb R^n,\mathbb R^n)$ s.t. $$\det J_f(x)\neq 0$$ for all $x\in\mathbb R^n$, why $f$ is injective ?

I tried many things, but I'm not allowed to conclude. For example, suppose $x\neq y$ and suppose $f(x)=f(y)$ Then $$f(x)=f(y)+J_f(y)(y-x)+(y-x)\varepsilon(x)$$ where $\lim_{x\to y}\varepsilon(x)=0$, and $$(y-x)\varepsilon(x)=\begin{pmatrix}(y_1-x_1)\varepsilon_1(x)\\ \vdots \\ (y_n-x_n)\varepsilon_n(x)\end{pmatrix}$$

thus $$J_f(y)(y-x)+(y-x)\varepsilon(x)=0.$$ Can I conclude that $J_f(y)=0$ and thus having a contradiction ?

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  • $\begingroup$ This is the condition to conclude that $f$ is locally injective using the inverse function theorem... but you cannot say anything about the global behavior of $f$. $\endgroup$ – pbn990 May 30 '18 at 17:41
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This is not true. A counterexample is given by $f:\Bbb C\to\Bbb C$, $f(z)=e^z$; since $f'(z)\ne0$ the function must have non-singular Jacobian when regarded as a map from $\Bbb R^2$ to itself.

Translating to a real-variables example: Define $F:\Bbb R^2\to\Bbb R^2$ by $$F(x,y)=(e^x\cos(y),e^x\sin(y)).$$

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