Let $f:\mathbb{R}^2\to\mathbb{R}$ be a $C^1$ function. Prove that the restriction is not injective. Let $f:\mathbb{R}^2\to\mathbb{R}$ be a $C^1$ function. And let $D$ an open subset of $\mathbb{R}^2$. Prove that the restriction of $f$ to $D$ is not injective.
Im trying to solve this but i dont know how...
The problem have a hint: Use the inverse theorem applied to an appropiate transformation in $\mathbb{R}^2$. 
 A: Assume by contradiction that $f_D$ is injective. 
Let $U \subset D$ be an open connected subset of $D$. Then $f|_U$ is also injective. As $U$ is connected then $f(U)$ is connected, hence an interval. Let $d \in f(U)$ be an interior point.
As $f$ is injective, there exists an unique $e \in U$ so that $f(e)=d$.
Then $f$ is a continuous bijection between the connected set $U \backslash \{e \}$ and the disconnected set $f(U)\backslash \{ d \}$. 
A: $f|_D$ cannot be injective. Now $f$ is not constant (because then it cannot be injective), so that $Df\neq 0$.Let $x$ be a point where $Df \neq 0$. Since $f$ is $C^1$, there exists a 'hood $U_x$ of $x$ where $Df(y) \neq 0 $. This means, by the inverse function theorem, that $f |_{U_x} \rightarrow f(U_x)$ is a diffeomorphism, so that there is a differentiable local inverse. But, by a dimension argument, using Sard's theorem, this is not possible: every point in $f(U_x) \subset \mathbb R$ is a critical point of $Df^{-1}|_{U_x}$, i.e., the image of $f^{-1}|_{U_x}$ has measure zero in $\mathbb R^2$ , so, in particular, $f^{-1}|_{U_x}$ cannot map into the open set $U_x$ , contradicting the existence of the local diffeomorphism.
A: Let $x \in D$, and $\epsilon>0$ such that $B(x,2 \epsilon) \subset D$.
Let $\gamma:[0,1] \to D$ be given by $\gamma(t) = x + \epsilon(\cos(2 \pi t), \sin(2 \pi t))$. We note that $\gamma(0) = \gamma(1)$, the restriction to $[0,1)$ is injective and since $[0,1]$ is compact and connected, $\gamma([0,1])$ is compact and connected.
Since $f$ is continuous, $f \circ \gamma([0,1])$ is also compact and connected, hence an interval. Let $a=\min_t f \circ \gamma(t)$, $b=\max_t f \circ \gamma(t)$. Then $f \circ \gamma([0,1]) = [a,b]$.
If $a=b$, we have $f(\gamma(0)) = f(\gamma(\frac{1}{2}))$, and since $\gamma(0)) \ne \gamma(\frac{1}{2})$, we see that $f$ is not injective.
Suppose $a<b$, then for some $t_b$ we have $b=f(\gamma(t_b)) $. Re-parametrize the curve $\gamma$ as $\eta(t) = \gamma((t+t_b) \mod 1)$, and note that $\eta$ is continuous and restriction to $[0,1)$ is injective.
Then we have $b=f(\eta(0))= f(\eta(1)$. Let $t^*$ be such that $a=f(\eta(t^*))$.
Then let $c= \frac{1}{2}(a+b)$, then by the intermediate value theorem, there exists some $t_1 \in (0,t^*)$ and $t_2 \in (t^*,1)$ such that $f(\eta(t_1)) = f(\eta(t_2)) = c$. Since $\eta(t_1) \ne \eta(t_2)$, we see that $f$ is not injective.
