Let \(f:\mathbb R^n\to\mathbb R^n\) be continuously differentiable. Assume its Jacobian matrix has rank \(n\) everywhere and that \(f\) is proper: \(f^{-1}(K)\) is compact whenever \(K\) is compact. Prove that \[ f(\mathbb R^n)=\mathbb R^n. \]
Solution
Put \[ U\coloneqq f(\RR^n). \]
1. The set \(U\) is open in \(\RR^n\).
At every \(x\in\RR^n\), the Jacobian matrix \(Df(x)\) has rank \(n\), hence is invertible. By the inverse function theorem, there is an open neighborhood \(V_x\) of \(x\) such that \[ f|_{V_x}:V_x\longrightarrow f(V_x) \] is a diffeomorphism onto an open subset \(f(V_x)\) of \(\RR^n\). Therefore \[ U=\bigcup_{x\in\RR^n}f(V_x) \] is open.
2. The set \(U\) is closed in \(\RR^n\).
Let \((y_j)\) be a sequence in \(U\) converging to \(y\in\RR^n\). Choose \(x_j\in\RR^n\) with \[ f(x_j)=y_j. \] The set \[ K\coloneqq\{y\}\cup\{y_j:j\ge1\} \] is compact because it consists of a convergent sequence together with its limit. Properness therefore makes \(f^{-1}(K)\) compact. Since every \(x_j\) belongs to \(f^{-1}(K)\), some subsequence satisfies \[ x_{j_\ell}\longrightarrow x\in f^{-1}(K). \] Continuity of \(f\) gives \[ f(x) = \lim_{\ell\to\infty}f(x_{j_\ell}) = \lim_{\ell\to\infty}y_{j_\ell} = y. \] Thus \(y\in U\). Hence \(U\) contains the limit of every convergent sequence from \(U\), and therefore \(U\) is closed in the metric space \(\RR^n\).
3. One has\[ \boxed{f(\RR^n)=\RR^n}. \]
The set \(U\) is nonempty, since \(\RR^n\) is nonempty. By steps 1 and 2, \(U\) is both open and closed in \(\RR^n\). Since \(\RR^n\) is connected, its only nonempty subset that is both open and closed is \(\RR^n\) itself. Thus \(U=\RR^n\).
4. Q.E.D.
Step 3 is the desired surjectivity statement.