A $C^{\infty}$ function from $\mathbb{R}^2$ to $\mathbb{R}$ Сould any one help me  how to show $C^{\infty}$ function from $\mathbb{R}^2$ to $\mathbb{R}$ can not be injective?

If f is constant then $f$ is trivially not injective.
Let $f$ be a non-constant. $C = \{\text{ critical points }\}$=$\{p: df(p) \text {is singular }\}$
We know that by Sard's Theorem $f(C)$ is of measure $0$. Hence
there is a regular point $p$. It's inverse image is either empty or
consists of some regular points.
Suppose for all regular $p$, $f^{-1}(p)$ were empty, then then $f(C)$ has
only critical values. Continuity means that $f(C)$ which has measure $0$,
should also be connected subset of $\mathbb{R}$. The only connected sets are
intervals. So $f(C)$ must be a point i.e., $f$ is a constant, a
contradiction. Therefore there is some regular value $p$ with a nonempty
pre-image, which has to be a $1$-manifold which cannot be a point.
Therefore f is not injective.
