I have a question on the famous paper Über die stetige Abbildung einer Linie auf ein Flächenstück (which translates roughly as On the continuous mapping of a line onto a square) by D. Hilbert. Let the line be $[0,1]$ and the square $[0,1]\times [0,1]$. Then Hilbert notes three points.
i) The mapping is continuous.
ii) Every point in the square corresponds to either one, two or four points on the line.
iii) "[...] sich leicht eine eindeutige und stetige Abbildung finden lässt, deren Umkehrung eine nirgends mehr als dreideutige ist" (sorry, I do not know how to translate that...)
Now i) and ii) I guess I have understood. As i) follows by the numbering scheme and the subdivision, which is such that if $\left|\frac{k}{4^N} - x\right| < \frac{1}{4^N}$ for $k \in \{0,\ldots, 4^N\}$, then $x$ gets mapped onto a point in a subsquare of area $1/4^N \cdot 1/4^N$, as in each step behind the $N$-th step, the points of the form $1/4^n, n > N$ of $[0,1]$ around $x$ all getting mapped into a subsquare where one endpoint is $(k/4^N, k/4^N)$. For ii), the point $(k/2^N, k/2^N)$ is approximated by four different sequences in $[0,1] \times [0,1]$ which correspond to four different point in $[0,1]$, for example $(1/2,1/2)$ by one converging to $1/4$, one to $2/4$ and one to $3/4$ and one to $1$, and similar if $(1/2^N, y)$ and $y$ is not a power of $1/2$ then the point corresponds to two points in $[0,1]$ and if $(x,y)$ where neither $x$ nor $y$ is a power of $1/2$ there is a unique sequence of subsquares enclosing this point, corresponding to a unique sequence in $[0,1]$.
But how to understand iii)? What D. Hilbert means by "dreideutig"? If a mapping is injective, then its inverse has to be injective too, which in german means "eindeutig", and "dreideutig" guess means something like "choosing among at most three points", but that does not make sense?
EDIT: A picture of the original construction taken from the mentioned paper:
