Under which conditions the domain of the surjective function $f:[a,b]\times[c,d]\to[0,1]^{2}$ can be split s.t. the restrictions are bijective? This is a follow-up question to this.
Since it is not always possible to construct such partition, I would like to know if there are additional restrictions which we could impose so that the wanted partition exists in the $n$-dimensional setting ($n\geq 1$).
 A: The trivial answer to this question is tautological: such a partition exists if and only if it exists.

A more informative answer is: almost never (unless such a partition obviously exists). To be more specific, consider the following counterexample, when such a partition does not exist. Define the function $f\colon[0,1]^2\to[0,1]^2$ by the formula
$$f(x,y):=\Big(x,\frac{y(y-g(x))^2}{(1-g(x))^2}\Big)$$
for $(x,y)\in[0,1]^2$, where $g\colon[0,1]\to[0,1/2]$ is (say) a smooth strictly increasing function. Then the function $f$ is smooth and surjective.
However, for any finite set of rectangles with sides parallel to the coordinate axes such that the union of these rectangles is $[0,1]^2$, the restriction of $f$ to at least one of these rectangles is not bijective. Indeed, since $g$ is strictly increasing, the graph $G:=\{(x,g(x))\colon x\in[0,1]\}$ of $g$ has only finitely many (actually, at most two) points on the boundary of any of those rectangles. Since we only have finitely many of those rectangles, there will be a point $(u,v)=(u,g(u))\in G$ that is in the interior of one of those rectangles, say rectangle $R$. On this rectangle $R$, the function $f$ will not be bijective -- because for any small enough $t>0$ the equation $f(x,y)=(u,t)$ will have at least two distinct solutions in $R$. Indeed, the equation $f(x,y)=(u,t)$ can be rewritten as the system of equations
$$x=u$$
and
$$t=\frac{y(y-v)^2}{(1-v)^2},\tag{1}$$
and for each small enough $t>0$ equation (1) has two distinct roots $y\approx v\pm(1-v)\sqrt{t/v}$, close to $v$.

What is the lesson of this example? In this example, we can refer to the graph $G$ of the function $g$  as the branching curve, near which the bijectivity cannot hold. The property of this branching curve that was used in the example is that the curve has only finitely many points on the boundary of any rectangle. So, unless a wanted partition obviously/manifestly exists, we will always have a branching curve that has only finitely many points on the boundary of any rectangle, and then the wanted kind of partition will not exist.
The above discussion concerns the dimension $n=2$. The case $n>2$ is similar -- in that case, we will have branching hypersurfaces instead of branching curves.

In the linked question, you said that the problem came up from your personal research. I'd guess that your research took a wrong direction, and that you may actually consider partitions not necessarily into rectangles.
