Show that if $f : \mathbb{R}^{2} \to \mathbb{R}$ continuously differetiable then $f$ is not inyective 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 
 A: Actually the differentiable hypothesis is unnecessary, we only need continuity of f. If f were injective then $f^{-1}(a)$ is a single point for all a in the image of f. However $\mathbb{R}$ - {a} is disconnected while $\mathbb{R}^2$ minus a point is connected. A fact from general topology says that the image of a connected set under a continuous mapping must be connected and so we have a contradiction.
Note: I am not sure if this is an answer you will find acceptable, but this question has been tagged as general topology so hopefully this is relevant to you.
P.S. A similar argument works for all dimensions.
A: Suppose that $f$ is injective. Then $f(0,0)\ne f(1,0)$.
On $[0,1]$, consider the functions $g(t)=f(t,t-t^2)$ and $h(t)=f(t,t^2-t)$. Note that $g(0)=h(0)=f(0,0)$ and $g(1)=h(1)=f(1,0)$
The Intermediate Value Theorem says that for some $t_g\in(0,1)$, $g(t_g)=\frac{f(0,0)+f(1,0)}2$ and for some $t_h\in(0,1)$, $h(t_h)=\frac{f(0,0)+f(1,0)}2$. This means that
$$
f(t_g,t_g-t_g^2)=g(t_g)=\frac{f(0,0)+f(1,0)}2=h(t_h)=f(t_h,t_h^2-t_h)
$$
However, $t_g-t_g^2\gt0\gt t_h^2-t_h$. Thus, we have $f(t_g,t_g-t_g^2)=f(t_h,t_h^2-t_h)$ yet $(t_g,t_g-t_g^2)\ne(t_h,t_h^2-t_h)$. Thus, $f$ is not injective.
A: I'd recommend the connectedness proof, but here's one using the implicit function theorem:
Suppose $(a,b) \in \mathbb{R}^2$ is not a critical point and set $c = f(a,b)$. 
Version 1: Then by the implicit function theorem there's some $\epsilon$ such that $f^{-1}(c) \cap B_\epsilon((a,b))$ is a manifold of dimension $2-1 = 1$. It's non-empty, so it contains more than one (in fact uncountably many) points.]
Version 2 (in more elementary language, using the wikipedia statement of the implicit function theorem): Then at least one partial derivative at $(a,b)$ is nonzero. WLOG let $\frac{\partial f}{\partial x}|_{(a,b)} \neq 0$. Then by the implicit function theorem there's an open set $U$ containing $a \in \mathbb{R}$ and an open set $V$ containing $b \in \mathbb{R}$ and a function $g: U \rightarrow V$ such that $f(x,g(x)) = 0$ for all $x \in U$. In particular, there are uncountably many points (one for every point of $U$, at the least) such that $f(x,y) = c$.
So if there's even one point in $\mathbb{R}^2$ which is not a critical point, then we're done.
What if all points in $\mathbb{R}^2$ are critical points? Then in particular all partial derivatives are zero, so by the mean value theorem (along lines $x = \textrm{constant}$ and along lines $y = \mathrm{constant}$) $f$ is constant. Therefore it isn't injective in this case either.

Here's a sketch of the generalization to $\mathbb{R}^3 \rightarrow \mathbb{R}^2$ (higher cases are the same idea as this one; I'll use this so the notation and details are easier):
We have $f: \mathbb{R}^3 \rightarrow \mathbb{R}^2$ continuously differentiable and want to show it's not injective.
Case 1: There exists a point in $\mathbb{R}^3$ at which the the Jacobian matrix (i.e. derivative matrix) of $f$ is rank two. In this case, the implicit function theorem can be applied as above.
Case 2: The derivative matrix of $f$ is everywhere rank zero. In this case, the function is constant, as above.
Case 3: There exists a point $(a,b,c) \in \mathbb{R}^3$ at which the derivative matrix of $f$ is rank one. WLOG $\frac{\partial f_1}{\partial x} \neq 0$ and $f(a,b,c) = (0,0)$. Consider $f_1: \mathbb{R}^3 \rightarrow \mathbb{R}$ the first coordinate. Then by the implicit function theorem we get a subset $U \subset \mathbb{R}^2$ and a map $g: U \rightarrow \mathbb{R}$ such that $f_1(x,y,g(x,y)) = 0$ for $(x,y) \in U$. Now consider $h: U \rightarrow \mathbb{R}$ given by $h(x,y) = f_2(x,y,g(x,y))$. Then if $h$ is not injective, then $f$ is not injective. But we know $h$ is not injective by the work we did for $\mathbb{R}^2 \rightarrow \mathbb{R}$ (it works for open sets in $\mathbb{R}^2$ just the same as for $\mathbb{R}^2$ itself).
A: 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 $f$.
