Why is this mapping a diffeomorphism? Let $\varphi:\mathbb{R}^2 \rightarrow \mathbb{R}$ satisfy $$\varphi_{xx}\varphi_{yy} - \varphi_{xy}^2 = 1$$.
Then define the mapping $(x,y) \mapsto (\xi, \eta)$ by:
$$
\xi(x,y) = x+\varphi_x(x,y)\:, \: \eta(x,y) = y+\varphi_y(x,y)
$$
It is claimed that this mapping is a diffeomorphism. I have a proof which involves some inequalities here but I would like to show this in a more "direct" way. I have calculated the Jacobian as:
$$
J = \begin{bmatrix}
\xi_x & \xi_y \\
\eta_x & \eta_y
\end{bmatrix}
=
\begin{bmatrix}
1+\varphi_{xx} & \varphi_{xy} \\
\varphi_{xy} & 1+\varphi_{yy}
\end{bmatrix}
$$
This has the determinant $$\det(J) = 1+\varphi_{xx} + \varphi_{yy} + \varphi_{xx}\varphi_{yy} - \varphi_{xy}^2$$
Using the hypothesis this is $\det(J) = 2+\varphi_{xx} + \varphi_{yy}$.
What I think: I cannot prove that the above is definitely not equal to $0$, other than assuming from the beginning that $\varphi_{xx}, \varphi_{yy} \geq 0$. EDIT: This is simply achieved by changing the sign of $\varphi$ in case it so happens that $\varphi_{xx}, \varphi_{yy} < 0$ (as suggested in this paper).
 A: I notice that $J$ has $\det(J) = \operatorname{tr}(J)$. The eigenvalues $\lambda_1,\lambda_2$ of $J$ must satisfy $\lambda_1\lambda_2 = \lambda_1 + \lambda_2$. If ever one is zero, the other must be as well. So if $J$ is singular, it must be the zero matrix.
If $J$ is the zero matrix, then $\varphi_{xx} = \varphi_{yy} = -1$ and $\varphi_{xy} = 0$. By integrating the latter equation we have that $\varphi(x,y) = u(x) + v(y)$. Integrating $u$ with respect to $x$ and $v$ with respect to $y$,
$$ \varphi(x,y) = cx + dy + e - \frac{1}{2}(x^2 + y^2) $$
Without additional hypotheses on $\varphi$, looks like $\varphi(x,y) = -\frac{1}{2}(x^2 + y^2)$ is a counterexample to the claim that $(x,y) \mapsto (x + \varphi_x, y + \varphi_y)$ is a local diffeomorphism.
More generally, this actually shows $(x,y)\mapsto (x+\varphi_x+y+\varphi_y)$ is a local diffeomorphism if and only if $\varphi$ is not of the form $-(x^2 + y^2)/2 + \mbox{linear}$.
If we require that $\varphi$ is convex, this rules out the only class of functions that makes $J$ singular, so we get that $(x,y)\mapsto (x+\varphi_x, y+\varphi_y)$ is a local diffeomorphism. I think the paper you linked needs to make convexity a hypothesis, as flipping the sign of $\varphi$ is clearly not WLOG.
