# 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?

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Try the implicit function theorem. –  user12477 Jun 21 '12 at 9:33
Your question is a particular case of this one –  Georges Elencwajg Jun 21 '12 at 10:57
May I apply this result to solve the problem? Here is the result: Assume $f:M\rightarrow N$ is a $\mathbb{C}^{\infty}$, let $n\in N$ and $P=f^{-1}(n)\neq\phi$ and $df:M_m\rightarrow N_{f(m)}$ is surjective $\forall m\in P$. Then $P$ has a unique manifold structure such that $(P,i)$ is a submanifold of $M$, where $i$ is inclusion map . Moreover $i:P\rightarrow M$ is an imbedding and the dimension of P is $c-d$ where $dimM=c$ and $dim N=d$ –  Bunuelian Trick Jun 22 '12 at 8:26
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## 4 Answers

$\mathbb{R}^2\setminus\{x\}$ is connected for any $x\in\mathbb{R}^2$.

Only continuity is required for the argument.

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could you explain a little more? –  Bunuelian Trick Jun 22 '12 at 7:04
@Mex: If the map $f:\mathbb{R}^2 \to\mathbb{R}$ were injective, then the restriction $f:\mathbb{R}^2\setminus \{x\} \to \mathbb{R}\setminus \{f(x)\}$ is well-defined and still continuous. –  Willie Wong Jun 22 '12 at 9:40
so what? where is the contradiction? –  Bunuelian Trick Jun 22 '12 at 11:49
@Mex: choose $x$ such that there exists $y,z$ with $f(y) < f(x) < f(z)$: this is always possible unless $f(\mathbb{R}^2)$ is contained in a two point set, in which case $f$ is not injective. Now using the fact that the image of a connected set under a continuous map is connected, you immediately obtain a contradiction. –  Willie Wong Jun 22 '12 at 13:30
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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 $\mathbb{R}$.

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so what is the relationship betwen injectivity and connectedness? –  Bunuelian Trick Jun 22 '12 at 7:05
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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.

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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.

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