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If $f: \mathbb{R}^2 \to \mathbb{R}$ is a $C^1$ function, then $f$ is not bijective.

Can someone explain why this is true or give me some hint? I don't realize how is bijection related with being a $C^1$ function, is it somehow related with the inverse function theorem?

la flaca
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    Actually, continuity is enough. Try to use the fact that $\Bbb{R}^2$ is connected, and that continuous maps send connected sets into connected sets. – Crostul Apr 20 '16 at 13:19
  • @Crostul : may be that is too advanced? –  Apr 20 '16 at 13:34
  • @cello May be, I still don't get it. I guess the problem should be with injectivity since $\mathbb{R}$ and $\mathbb{R}^2$ have the same cardinality so surjectivity should not be a problem, right? – la flaca Apr 20 '16 at 13:56
  • @Eliana Find a partition of $\mathbb{R}^2$ into uncountably many connected subsets. Then think why this makes contradiction. – cjackal Apr 20 '16 at 14:39

3 Answers3

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Here is a solution requiring nothing more than basic calculus showing that continuity is enough.

Suppose $f: \Bbb{R}^2 \to \Bbb{R}$ is continuous. I show you by contradiction that $f$ is not injective. Suppose by contradiction that $f$ is injective.

Consider the three maps $g_1, g_2, g_3 : [0,1] \to \Bbb{R}^2$ defined by $$g_1(t) = (0,t)$$ $$g_2(t) = (t,0)$$ $$g_3(t) = (t,t)$$ It is obvious that these are continuous and injective. Consider the three (continuous and injective) compositions $f \circ g_1, f \circ g_2, f \circ g_3$ defined over $[0,1]$ with values in $\Bbb{R}$.

Then $$f(g_1([0,1])), f(g_2([0,1])), f(g_3([0,1]))$$ are three bounded closed intervals containing the number $z= f(0,0)$. Note that:

  1. Since $f \circ g_i$ is injective, $f(g_i([0,1]))$ does not consist of a single point. So $f(g_i([0,1])) = [a_i, b_i]$ with $a_i < b_i$.

  2. But $\{ z \}= [a_1, b_1] \cap [ a_2, b_2] = [a_1, b_1] \cap [ a_3, b_3] =[a_2, b_2] \cap [ a_3, b_3]$ by injectivity. And this is impossible.

Crostul
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Another proof for $f$ merely continuous: Because $\mathbb R^2$ is connected, so is $f(\mathbb R^2)$ by the continuity of $f.$ Thus $f(\mathbb R^2)$ is an interval. Choose a point $a$ in the interior of this interval. Then there is a unique $(b,c)\in \mathbb R^2$ such that $f(b,c)=a.$ But $\mathbb R^2 \setminus \{(b,c)\}$ is connected (because it's path connected), while $f(\mathbb R^2)\setminus \{a\}$ is not. Since $f$ preserves connectness, we have a contradiction.

zhw.
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As you can see in the other answers, continuity is already enough to derive a contradiction. In addition, here is a proof that uses that $f$ is continously differentiable:

Assume that $f$ is bijective and of class $C^1$. Write $f: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$, and consider its differential $d_{(x,y)}f = (f_x, f_y)$. Because $f$ is bijective, there is some point $(x_0, y_0)$ such that $d_{(x_0,y_0)}f \neq (a,0)$, where $a \in \mathbb{R}$ is arbitrary. Let $f(x_0,y_0) = z_0$ and consider

$F(x,y) = f(x_0,y_0) - z_0$

Then we can apply the implicit function theorem (since the second component of the differential at $(x_0,y_0)$ is not zero and thus invertible), getting open neighbourhoods $U,V \subseteq \mathbb{R}$ with $x_0 \in U, y_0 \in V$ and a function $g: U \rightarrow V$ with $g(x_0) = y_0$, such that

$F(x,y) = 0\, \Leftrightarrow\, y = g(x)$

on $U \times V$ or, in other words,

$f(x,g(x)) - z_0 \equiv 0\, \Leftrightarrow\, f(x,g(x)) \equiv z_0$

on $U \times V$, which is a contradiction to the bijectivity of $f$.

For your interest: there are (non-continuous) bijective maps from $\mathbb{R^2}$ to $\mathbb{R}$ (look for example here: Examples of bijective map from $\mathbb{R}^3\rightarrow \mathbb{R}$)

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