# Show $f$ is not $1-1$

Let $f:\Bbb R^2\to \Bbb R$ be a continuously differentiable function. Show that $f$ is not $1-1$.

I know I will need to use the Inverse Function Theorem and consider some open set A with $g:A\to \Bbb R^2$ defined by $g(x,y)=\big(f(x,y),y\big)$.

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Being continuous, $f$ assumes its min and max on the compact $S^1$. By the IVT, any value between min and max ias assumed once in each of the two arcs bounded by a minimal and a maximal point.

Correct me if I’m wrong, but I thought this was a more general result:

No function defined on a compact subset of $\mathbb{R}^2$ can be 1-1, basically by the pigeonhole principle.

The nonexistence of a utility function representing lexicographic preferences basically boils down to this principle.