IMC 2002 · Day 1 · P5

a. It does not exist. For each $y$ the set $\{ x : y = f(x) \}$ is either empty or consists of 1 point or is an interval. These sets are pairwise disjoint, so there are at most countably many of the third type.

b. Let $f$ be such a map. Then for each value $y$ of this map there is an $x_0$ such that $y = f(x)$ and $f'(x) = 0$, because an uncountable set $\{ x : y = f(x) \}$ contains an accumulation point $x_0$ and clearly $f'(x_0) = 0$. For every $\varepsilon > 0$ and every $x_0$ such that $f'(x_0) = 0$ there exists an open interval $I_{x_0}$ such that if $x \in I_{x_0}$ then $|f'(x)| < \varepsilon$. The union of all these intervals $I_{x_0}$ may be written as a union of pairwise disjoint open intervals $J_n$. The image of each $J_n$ is an interval (or a point) of length $< \varepsilon \cdot \operatorname{length}(J_n)$ due to Lagrange Mean Value Theorem. Thus the image of the interval $[0,1]$ may be covered with the intervals such that the sum of their lengths is $\varepsilon \cdot 1 = \varepsilon$. This is not possible for $\varepsilon < 1$.

Remarks. 1. The proof of part b is essentially the proof of the easy part of A. Sard's theorem about measure of the set of critical values of a smooth map.

2. If only continuity is required, there exists such a function, e.g.\ the first co-ordinate of the very well known Peano curve which is a continuous map from an interval onto a square.