I found this in Partial Differential Equations, Evans, page 627.
I would to like understand why Evans claims that $\det \Phi =\det \Psi=1$. Here, $y_i=x_i:= \Phi^{i}$, $i=1,\ldots,n-1$ and $y_n=x_n -\gamma(x_1, \ldots,x_{n-1})$ ($y=\Phi(x)$).
The inverse transform is similarly defined: $x_i=y_i$ when $i=1,\ldots,n-1$, $x_n=y_n + \gamma(x_1, \ldots,x_{n-1}) $.($x=\Psi(y)$). The function $\gamma :\mathbb R^{n-1}\to \mathbb R$ is of class $C^2$. The book says $\det \Phi =\det \Psi=1$.
I tried with $n=2$ as follows: The transformation $T$ defined on $\mathbb R^2$ by $y_1=x_1$ and $y_2=x_2 -\gamma(x_1)$. I chose the basis $e_1,e_2$ of $R^2$ so $[T(e_1)]_{e_1,e_2}=(1,-\gamma(1))$ and $[T(e_2)]_{e_1,e_2}=(0,1-\gamma(0))$, so when I do the determinant of the matrix that represents $T$ I obtain $1-\gamma(0)$, which is different from $1$.
Please, can somebody help me or give me some hint? Thank you. Here is a poor quality screenshot (sorry): https://i.stack.imgur.com/tzRqf.jpg
Its not clear what the determinant of $\Phi(x)\in\mathbb R^n$ should be. Instead it should be that the determinant of $\nabla\Phi(x) \in \mathbb R^{n\times n}$ is $1$. This has been corrected in the Second Edition:
The reason the determinant of $\nabla\Phi$ is identically one is because the matrix is lower triangular with all diagonal entries equal to 1. I write it out below in the special case of dimension 2: $$ \nabla\Phi(x) = \begin{pmatrix}\partial \Phi_1/\partial x_1 & \partial \Phi_1/\partial x_2 \\ \partial \Phi_2/\partial x_1 & \partial \Phi_2/\partial x_2 \end{pmatrix}(x) = \begin{pmatrix} 1 & 0 \\ -\gamma'(x_1) & 1\end{pmatrix}.$$
The determinant of the gradient of the inverse map follows.