$f: \mathbb{R}^2 \to \mathbb{R}$ is a continuously differentiable function (of class $C^1$). Show that mapping f can not be one-to-one mapping.

Question

$f: \mathbb{R}^2 \to \mathbb{R}$ is a continuously differentiable function (of class $C^1$). Show that mapping $f$ cannot be one-to-one mapping.

Let $D_1F(x,y) \neq 0$ for all $(x,y)$ for some open subset A and consider the function $g: A \to \mathbb{R}^2$ defined by the equality $$g(x,y) = (f(x,y),y)$$

Not sure if I'm doing it right so far, but how would I generalize it to $C^1$ mappings $f: \mathbb{R}^n \to \mathbb{R}^m$, $n > m$?

Answer 1

There are circles in ${\mathbb R}^n$ for $n > 1$, but no one-to-one images of circles in $\mathbb R$: any connected set in $\mathbb R$ can be disconnected by removing one point, a circle requires two.

Answer 2

for $r \in [0,1]$ define $$U_{r} = f(\{r\}\times(0,1))$$ if $f$ be one-to-one then $f$ over $\{r\}\times(0,1)$ is one-to-one for every $r$ and $U_{r}$ will be an open set in $\mathbb{R}$. now $U_{r}$s are uncountable distinct open sets in $\mathbb{R}$ and this one is'nt possible to take uncountable distinct opens in $\mathbb{R}$.

Answer 3

I think this should work:restrict the map to a compact subset K with non-empty interior of $\mathbb R^2$. Then $f|_K$ is a continuous bijection between compact and Hausdorff , so it is a homeomorphism. Choose an open subset of K; its image will be open in $\mathbb R $, which cannot happen, e.g., by invariance of domain. For the second case, you should have $n>m$, or the injection $(a_1,..,a_n) \rightarrow (a_1,..,a_n,0,0,..,0)$ will be a counterexample .