Is the Inverse of the Vectorised Solid Angle Equation for $n$ Circular Discs Continuous? I have a continuous function$^{*1}$ that takes in 3 arguments, and returns 24 outputs. I want to know if the inverse of this function is continuous.
The 3 input arguments are the x, y, and z position of a particle in 3D space within an array of circular discs, and the 24 outputs are the solid angles subtended by the 24 discs, as shown in the picture bellow.

(This picture shows 18 detectors, the 24 are in a similar arrangement: 2 layers of 4$\times$3)
$$$$
The equation for the solid angle for each disk is given by this paper. The equations are shown bellow:
\begin{align} \Omega &= 2\pi - \frac{2L}{R_{max}}K(k) - \pi\Lambda_{0}(\xi , k) &for\;\; r_0 > r_m\label{eq:elliptic1}\\[10pt] &= \pi - \frac{2L}{R_{max}} K(k) &for\;\; r_0 = r_m\label{eq:elliptic2}\\[10pt] &= - \frac{2L}{R_{max}} K(k) + \pi\Lambda_{0}(\xi , k) &for\;\; r_0 < r_m\label{eq:elliptic3} \end{align}
I've inputted 2000 random particle locations and then used the results from these equations to train a Neural Network.  This Neural Network was trained using the 24 solid angles as its inputs, and the position of the particle in 3D space as its outputs.
This means the Neural Network approximates a function that takes in 24 inputs, and returns 3 outputs (x, y, and z positions).
What I want to know is: Is the function that is being approximated a continuous function?
Is there even a way to know this for certainty? Any answer or comment is very much appreciated.
Thanks!

$^{*1}$ Only continuous if the particle is within the array of discs. This can be taken to be always true for the application I am using this for.
 A: For starters, let's consider a single disk $D$ and its solid angle value, $\theta$. Look at the locus of points in 3D that give this angle value with this disk. This locus is some sort of surface $S(D,\theta)$ in 3D space. It might even be a sphere, actually. Probably not. But it doesn't matter.
Now take three such disks $D_1, D_2, D_3$ and their solid angles $\theta_1, \theta_2, \theta_3$. In order to satisfy our conditions, a point $P$ must lie on all three of the surfaces $S(D_1, \theta_1)$, $S(D_2, \theta_2)$, $S(D_3, \theta_3)$. In other words, the point $P$ must be a point of intersection of the three surfaces. Generally, you would expect 3 surfaces in 3D to intersect in only a finite number of points, unless they have some special dependency relationship, so there are only a finite number of possibilities for $P$.
Now add a fourth disk. To me, it seems highly unlikely that its locus surface would pass through any of the points of intersection of the first three locus surfaces. 
Saying it more briefly ... it seems highly unlikely that the 24 surfaces $S(D_i,\theta_i)$ will have any common points of intersection. 
