Let $f: \mathbb R^2 \to \mathbb R^2$ be $C^1$, $D_f(x)$ is invertible everywhere, and $\lim_{|x| \to \infty}|f(x)| = \infty$
Show that $\min_{x \in \mathbb R^2}|f(x)-a|$ exists, and that $f$ is onto.
I realize that $f$ is locally invertible everywhere but I'm not sure how to use it to prove what we want.
Let $g(x) = |f(x)-a|^2$
See that $\lim_{|x| \to \infty}g(x) = \lim_{|x| \to \infty}|f(x)-a|^2 \geq\lim_{x \to \infty}(|f(x)| - |a|)^2 = \infty$
Let $B(0, R)\subset \mathbb R^2$ be a closed ball around $0$ such that $g(x) > g(0)$ outside the ball. We know such a ball exists because $g$ is non negative and $\lim_{|x| \to \infty}g(x) = \infty$
This is a compact set and $g$ is continuous and so $g$ attains it's minimum somewhere in this set. It can't attain the minimum on the boundary of the ball because of continuity, and so we must have a local minimum somewhere inside.
To sum up, there is a point $x_0 \in (B(0, R))^\circ$ such that $\nabla g(x_0) = 0$.
But $\nabla g(x_0) = 2(f(x_0)-a)D_f(x_0) = 0$ but $D_f(x_0)$ is invertible, so it follows that $f(x_0) = a$. So the minimum exists, it is zero, and $f$ is onto.