TITLE: Let $f : \mathbb{R}^{2} \to \mathbb{R}$ be of class $C^{1}$. Show that $f$ not is injective.
QUESTION [6 upvotes]: Let $f : \mathbb{R}^{2} \to \mathbb{R}$ be of class $C^{1}$. Show that $f$ not is injective. 
Well, I know that $f$ is injective when $f(x) = f(y)$ implies $x = y$. Since $f$ is of class $C^{1}$ then $f$ is differentiable in $\mathbb{R}^{2}$ such that its derivative is continuous. I can not figure out the ideas. This question seems to suggest a counter example, but none is working. Thanks for the help!

REPLY [2 votes]: In fact, all you need is that $f$ is continuous.  Consider two paths from $(-1,0)$ to $(1,0)$, say the top and bottom halves of the unit circle.  By the intermediate value theorem, for any $y$ between $f(-1,0)$ and $f(1,0)$ there is a point on each path where $f$ has the value $y$.  Therefore $f$ can't be injective.