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?
 A: $\mathbb{R}^2\setminus\{x\}$ is connected for any $x\in\mathbb{R}^2$.
Only continuity is required for the argument.
A: If we remove three points from the domain it will be connected. In $\mathbb{R}$ the connected sets are intervals, so if we remove three point from an interval it will be disconnected. So there can not exist a continuous injective function from $\mathbb{R}^2$ to $\mathbb{R}$.
A: 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.
A: Suppose that $f:\mathbb R^2\rightarrow\mathbb R$ is injective and continuous. 
Then $f$ induces a function $\tilde f:\mathbb R^2\rightarrow \mathbb R\hookrightarrow\mathbb R^2$ which is injective and continuous given by $\tilde f(x)=(f(x),0)$.
By the invariance of domain, $\tilde f(\mathbb R^2)$ is a non empty open set of $\mathbb R^2$. But $\tilde f(\mathbb R^2)\subset\mathbb R\times\{0\}$. Contradiction.
