TITLE: Show that if $f : \mathbb{R}^{2} \to \mathbb{R}$ continuously differetiable then $f$ is not inyective
QUESTION [7 upvotes]: Well my question this time is:
How to show that $f : \mathbb{R}^{2} \to \mathbb{R}$ continuously differetiable then $f$ is not inyective 
I was trying to consider the function $g(x,y)=(f(x,y),y)$, but I don't know how to proceed with $g$, I know I have to use the implicit function theorem but when I was thinking of it, I remember that nearly $(x_0,y_0)$, $f$ must be zero, well $f(x_0,y_0)=0$, then the only tool I thought it could help I can't apply it.
Can someone help me please with this problem?, and once you have it for this case, Can we generalize to $R^{m}$ and $R^{n}$?(of course $m<n$) 
Thanks a lot in advance
My implicit function theorem:
Let $A \subset R^{m} \times R^{n}$ open, $(x_0,y_0) \in A$, $f:A \to R^{m}$ continously differentiable nearly $(x_0,y_0)$ and  $f(x_0,y_0)=0$. Let M be the matrix of $m \times m$ given by:
$$D_{n+j}f^{i}(x_0,y_0)$$
and suppose that $\det(M) \not= 0$, then there exists an open set $U \subset R^{n}$ $x_0 \in U$, and an open $V \subset R^{m}$, $y_0 \in V $, such that, for each $x$ there exists a unique $g(x) \in V$ such that $f(x,g(x))=0$, $g$ differentiable
I don't know Topology, or more than introduction to analysis and a little bit of multivariable analysis, thanks a lot for your help

REPLY [2 votes]: By contradiction, suppose that f is one-to-one. Then f is not constant and at least one of the next conditions is true: $D_{1}f(x,y)\neq 0$ or $D_{2}f(x,y)\neq 0$ for some $(x,y)\in R^{2}$. 
WLOG suppose $D_{1}f(x,y)\neq 0$ (the case $D_{2}f(x,y)\neq 0$ is analogous) by the continuity of $D_{1}f(x,y)$ there is an open set $A$ that contains $(x,y)$ where $D_{1}f(x,y)\neq 0$ for all $(x,y)\in A$. Define $g:A\rightarrow R^2$ by $g(x,y)=(f(x,y),y)$ then $det[g'(x,y)]=D_{1}f(x,y)\neq 0$. This implies that $g(A)$ is open and by the inverse function theorem $g$ has a differentiable inverse in the open set $g(A)={(f(x,y),y):(x,y)\in A}$. 
However, this is not possible because for some $(x_{0},y_{0})\in A$ put $f(x_{0},y_{0})=b$, then $g(x_0,y_0)=(b,y)$ and the function $g^{-1}$ is not defined in any point $(b,z)$ because in that case $f(x_0,z)=b$ which contradicts the inyectivite of $f$.