Is it always possible to partition $[a,b]\times[c,d]$ into disjoint blocks $D_{ij}$ s.t. $\left.f\right|_{D_{ij}}$ is bijective? Consider the function given by $f:[a,b]\times[c,d]\to[0,1]^{2}$ such that $0\leq a < b \leq 1$, $0 \leq c < d \leq 1$.
Moreover, we do also have that $f\in C^{1}([a,b]\times[c,d],[0,1]^{2})$ and it is surjective.
My question
Is it always possible to partition $[a,b]\times[c,d]$ into disjoint blocks $D_{ij} = [x_{i},x_{i+1}]\times[y_{j},y_{j+1}]$, where $1\leq i \leq m$ and $1\leq j\leq n$, such that $\left.f\right|_{D_{ij}}$ is bijective?
If so, is there a minimum number of blocks $D_{ij}$ that satisfy this restriction?
Here I assume the function $f$ is not constant anywhere and $|f^{-1}(\{(x,y)\})| < N$ for every $(x,y)\in[0,1]^{2}$.
Such question is not homework. It came up from my personal research.
If this question is not adequate to this site, please let me know.
EDIT
The follow-up question to this is addressed here.
 A: The answer is no.
E.g., let $[a,b]=[c,d]=[0,1]$ and
$$f(x,y):=(g(x),y)$$
for $(x,y)\in[0,1]^2$, where
$$g(x):=c\,h(x),$$
$$h(x):=x^p (1+a \sin\ln x)$$
for $x\in(0,1]$ with $h(0):=0$,
$$p\in(1,\infty),\quad1>a>\frac p{\sqrt{p^2+1}},\tag{0}$$
and $c:=1/\max_{x\in[0,1]}h(x)$. Then $f$ is a surjective $C^1$ map from $[0,1]^2$ to $[0,1]^2$.
Also, for any $(x,y)\in[0,1]^2$, any $u\in(0,1]$, and any $v\in[0,1]$ the equality $f(x,y)=(u,v)$ implies $y=v$ and
$$\Big(\frac{u/c}{1+a}\Big)^{1/p}\le x\le\Big(\frac{u/c}{1-a}\Big)^{1/p}$$
and hence
$$\frac{\ln(u/c)}p-\frac{\ln(1+a)}p\le \ln x\le\frac{\ln(u/c)}p-\frac{\ln(1-a)}p,\tag{1}$$
so that $\ln x$ varies by at most $\frac{\ln(1+a)}p-\frac{\ln(1-a)}p=O(1)$ uniformly in $u\in(0,1]$.
Also,
$$g'(x)=cx^{p-1} [p+a (p \sin\ln x+\cos\ln x)] \\
=cx^{p-1} [p+a\sqrt{p^2+1}\,\sin(t+\ln x)]\tag{2}$$
for some real $t$ (depending only on $p$ and $a$) and all $x\in(0,1]$.
So, given the condition (1), $g'(x)$ can change the sign no more than $n$ times, for some natural $n$ depending only on $p$ and $a$. Therefore, $|f^{-1}(u,v)|\le n+1$ for any $(u,v)\in(0,1]\times[0,1]$. Also, $f^{-1}(0,v)=\{(0,v)\}$ for any $v\in[0,1]$. So, $|f^{-1}(u,v)|\le n+1$ for any $(u,v)\in[0,1]\times[0,1]$.
On the other hand, it follows from (2) and (0) that $g'$ changes the sign infinitely many times in any right neighborhood of $0$. Therefore, the restriction of $f$ to any rectangle with a vertex at $(0,0)$ is not bijective.

For an illustration, below are the graphs $\{(e^{-1/t},\ln h(e^{-1/t}))\colon0<t<1\}$ (left) and $\{(e^{-1/t},\ln h(e^{-1/t}))\colon0<t<0.1\}$ (right) for $p=3/2$ and $a=9/10$. These graphs are non-linearly rescaled (horizontally and vertically, for better perception) versions of a graph of the function $h$ in a right neighborhood of $0$.

