Question
Suppose $f:\mathbb R^n\to \mathbb R^n$ is a $C^1$ function and $Df(x)$ is invertible for all $x\in \mathbb R^n$. Then $f$ is onto if $f^{-1}(K)$ is compact for all compact set $K\subseteq \mathbb R^n$.
I know that $f$ by inverse function theorem is locally invertible at each point but I don't know how to proceed.
Edit: By the hint from @Aphelli, we can prove that $\text{im} f$ is a clopen set. Notice that since $Df(x)$ is invertible for all $x\in \mathbb R^n$, by inverse function theorem, there exists open set $V_x, W_x\subseteq \mathbb R^n$ with $x\in V_x$, $f(x)\in W_x$ such that $f|_{V_x}:V_x\to W_x$ is invertible and its inverse is also $C^1$.
Note that $$\text{im}f=f\left(\bigcup_{x\in \mathbb R^n}V_x\right)=\bigcup_{x\in \mathbb R^n}W_x$$ and therefore $\text{im }f$ is open.
Let $y_n\in \text{im }f$ with $y_n\to y$. Then $\text{im} f$ is closed if $y\in \text{im}f$. Consider $K=\{y_n\}\cup \{y\}$, then $K$ is compact and therefore $f^{-1}(K)$ is also compact. Let $x_n\in f^{-1}(K)$ such that $f(x_n)=y_n$. Since $f^{-1}(K)$ is compact, there is a convergent subsequence $x_{n_k}$ such that $x_{n_k}\to x\in f^{-1}(K)$. Hence, $f(x_{n_k})=y_{n_k}$. Take $k\to \infty$, we have $f(x)=y\in \text{im}f$.