I am trying to find the inverse of a two dimensional map $f\left(\begin{bmatrix} x\\y\end{bmatrix}\right)$, For example $$ f\left(\begin{bmatrix}x\\y \end{bmatrix}\right) = \begin{cases} ax + by, &(x,y)\in D_1 \\ ax^3 + b x^4, & (x,y)\in D_2\end{cases} $$

For this question we have $f\left(\begin{bmatrix}x\\y\end{bmatrix}\right)$ is defined as

$$f\left(\begin{bmatrix} x\\ y\\ \end{bmatrix}\right) = \begin{cases} F_{0}\left(\begin{bmatrix} x\\ y\\ \end{bmatrix}\right), &y<a \\ \big(1-s(y)\big)F_{0}\left(\begin{bmatrix} x\\ y\\ \end{bmatrix}\right) + s(y) F_{1}\left(\begin{bmatrix} x\\ y\\ \end{bmatrix}\right), &a<y<b \\ F_{1}\left(\begin{bmatrix}x\\y \end{bmatrix}\right), & y>b\end{cases}$$

We know the inverses of $F_{0}$ and $F_{1}$.

The inverse of the map $f$ depends on the point $(x,y)$ we give, like for $y<a$, we will take the inverse of $f$ as the inverse of the map $F_{0}$ and if the $y$ value is greater than $b$, then we take the inverse of the map $T_{1}$. But when the $y$ value lies in between $a$ and $b$, then we take the inverse of $$\big(1-s(y)\big)F_{0}\left(\begin{bmatrix} x\\ y\\ \end{bmatrix}\right) + s(y) F_{1}\left(\begin{bmatrix} x\\ y\\ \end{bmatrix}\right),$$ which I am not sure how to calculate the inverse of seems like applying $(g.h)^{-1} = h^{-1}.g^{-1}$ will help in this case, still how to find the inverse of the piece o fthe function defined in $a<y<b$?

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    $\begingroup$ $s(y)F_0(v)$ is not a composite function. It is just a product. The rule you have mentioned is true for composite functions. I don't think products follow that rule. $\endgroup$ – Balakrishnan Rajan Mar 28 at 2:14
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    $\begingroup$ This function doesn't have an inverse if $F_0(x,y) = F_1(u,v)$ for some $(x,y)$ with $y<a$ and $(u,v)$ with $v > b$. $\endgroup$ – amsmath Mar 28 at 3:13
  • $\begingroup$ @BalakrishnanRajan yup you are right! $\endgroup$ – BAYMAX Mar 28 at 4:09
  • $\begingroup$ @amsmath nice, but let us suppose this doesnot happen! $\endgroup$ – BAYMAX Mar 28 at 4:09

A linear combination of invertible functions is not necessarily invertible. A very simple example is $f_1(x) = x$ and $f_2(x) = -x$. They are each invertible, actually being their own inverses, but $f(x) = f_1(x) + f_2(x) = 0$ is not invertible.

In your case, invertibility will depend on what $$\big(1-s(y)\big)F_{0}\left(\begin{bmatrix} x\\ y\\ \end{bmatrix}\right) + s(y) F_{1}\left(\begin{bmatrix} x\\ y\\ \end{bmatrix}\right)$$ results in, but even if this is invertible, you're not likely to find any easy formula to determine what it is from the base parts of $s$, $F_0$ and $F_1$.


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