Is each space filling curve everywhere self-intersecting?

Question

Consider a continuous surjection $f:[0,1]\to [0,1]\times[0,1]$. Is $$\{x:\exists(t_1\not=t_2) f(t_1)=f(t_2)=x\}=[0,1]\times[0,1]?$$

Answer 1

No. Consider for instance the Hilbert space-filling curve, on which Brian Hayes wrote a nice popular article recently (HTML, PDF).

This mapping from $[0, 1]$ to $[0, 1] \times [0, 1]$ can be viewed as a mapping from digit sequences $0.d_1d_2d_3\dots$ in base-$4$ (where $0 \le d_i < 4$) to $[0, 1] \times [0, 1]$. The first digit tells us which quadrant a point lies in, the second digit tells us which sub-quadrant of that it lies in, and so on. (See the second image on page 4.) Conversely, given a point in $[0, 1] \times [0, 1]$, we can invert the curve -- find all $t$ in $[0, 1]$ that are mapped to this point -- by noting down which quadrant it lies in, then which sub-quadrant of that it lies in, and so on, recursively.

The only points that are the image of multiple $t$ are those that, at some granularity, lie in multiple quadrants. These are precisely those points $(a, b)$ such that either $a$ or $b$ can be written as $\dfrac{m}{2^k}$ for some integers $k \ge 1$ and $1 \le m < 2^k$. Every other point is the image of exactly one $t$.

So for the Hilbert curve at least, the set of points at which the curve is "self-intersecting" in your sense is, far from being every point, actually of measure $0$.