I would like your help to simplify the construction of the set of solutions of a system of two equations. Answering the question requires knowing the definition of cumulative distribution function (hereafter CDF).
In short a CDF is a function $F$
1) from $\mathbb{R}$ to $[0,1]$
2) weakly increasing
3) $\lim_{x\rightarrow \infty}F(x)=1$, $\lim_{x\rightarrow -\infty}F(x)=0$
[see here for a more formal definition of CDF]
The system of equations
Consider the following system of equations $$ (\star)\hspace{1cm} \begin{cases} a=F(x)\\ b=F(y) \end{cases} $$ where
$a,b$ are known parameters both in $[0,1]$
$F,x,y$ are the unknowns with $x\in \mathbb{R}$, $y\in \mathbb{R}$, $F\in \mathcal{F}$ and $\mathcal{F}\equiv \{F:\mathbb{R}\rightarrow [0,1] \text{ s.t. } F \text{ is a continuous CDF}\}$
Fact 1 (not used)
For any $F\in \mathcal{F}$ there exists $(x,y)$ satisfying $(\star)$. Specifically, $x=Q_F(a), y=Q_F(b)$ solve the system where $Q_F$ is the quantile function associated with $F$.
[the quantile function is basically the inverse of the CDF, see here for a definition]
Fact 2
If $(F,x,y)$ solves $(\star)$ then $(F^{\mu},x+\mu,y+\mu)$ solves $(\star)$ too, where $F^{\mu}$ is obtained by "relocating" $F$ according to $\mu$, $\forall \mu \in \mathbb{R}$.
Fact 3
If $(F,x,y)$ solves $(\star)$ then $(F_{\sigma},\sigma x,\sigma y)$ solves $(\star)$ too, where $F_{\sigma}$ is obtained by "rescaling" $F$ according to $\sigma$, $\forall \sigma>0$.
Clarification on "relocating" and "rescaling"
I'm very naive in the way I think about relocating and rescaling.
Suppose that $F$ is the standard normal CDF. Then, relocating $F$ according to $\mu$ is intended as considering the CDF of the normal with mean $\mu$ and variance $1$ (if you look at the graphs here you see that the plot of the CDF is horizontally shifted by $\mu$). Rescaling $F$ according to $\sigma$ is intended as considering the CDF of the normal with mean $0$ and variance $\sigma^2$ (if you look at the graphs here you see that the plot of the CDF is rescaled by $\sigma$).
I guess that something similar can be always done for any (not necessarily normal) CDF $F\in \mathcal{F}$. Please advise me on this if I'm wrong.
What I want to do
Imagine that I have an algorithm searching for the set of solutions of $(\star)$.
When I run such an algorithm, for each group of solutions equivalent up to a "relocation" and up to a "rescaling", I want to save only one solution from the group.
My idea to achieve what I want to do
Instead of solving $(\star)$, solve wrto $(F,x,y)\in \mathcal{F}\times \mathbb{R}\times \mathbb{R}$ $$ (\star \star)\hspace{1cm} \begin{cases} a=F(x)\\ b=F(y) \end{cases} \text{ s.t. $x=0$ and $y\in \{-1,1\}$} $$ Attempted proof:
Notice that, by Fact 2, if $(F,x,y)$ solves $(\star)$ then $(F^{-x},0,y-x)$ solves $(\star)$ too.
Notice that, by Fact 3, if $(F,x,y)$ solves $(\star)$ then $(F_{|y|},\frac{x}{|y|}, \frac{y}{|y|})$ solves $(\star)$ too.
Hence, by solving $(\star \star)$, we achieve my objective.
Question: is my idea correct? Is it "without loss of generality"?
Further terminology
For readers familiar with econometrics, the procedure above is related to "scale and location normalisations in discrete choice models".