How do I reverse this function? F(a, b, c, d) = (ac, ad, bc, bd) Suppose I have a function that operates on 4 real numbers a, b, c, d in this way:

F(a, b, c, d) = (ac, ad, bc, bd)

Is there such a function F' that reverses the effect of this function? In other words, a function whose output I can plug into F and receive the original input. If so, how is it defined?

F'(a, b, c, d) = (?, ?, ?, ?)

Follow up: does F' have exactly one output for all possible inputs?
 A: No, there’s no inverse function. To see this, notice that $$F(1,1,1,1)=F(2,2,\frac{1}{2},\frac{1}{2})=(1,1,1,1).$$
A: One could write it in a matrix-notation to see the ambiguity for the inverse function, say $G()$. First we define:
$$ F(a,b,c,d)= \begin{matrix} 
  &&                \begin{bmatrix} c & d \end{bmatrix}\\
  &\times & \\
  \begin{bmatrix} a \\ b \end{bmatrix} 
     &=& \begin{bmatrix} ac & ad \\ bc & bd \end{bmatrix}
 \end{matrix} \implies (ac,bc,ad,bd)
$$
(Btw. we can see, that the determinant of the involved $4\times 4$-matrix is zero, so the inverse process should be indeterminate at least) 
The inverse function should reproduce from the $4\times 4$ matrix the two $1\times 2$ matrices, but this is underdetermined. We can easily assume $a=1$ to find a valid solution
$$ G(ac,bc,ad,bd)=
 \begin{matrix}
        & &\begin{bmatrix} ac & ad \end{bmatrix} \\ 
     & \times & \\
   \begin{bmatrix} 1 \\ b/a \end{bmatrix}    & =  &\begin{bmatrix}  ac & ad \\ bc & bd \end{bmatrix} 
 \end{matrix} \implies (1,b/a,ac,ad)
$$
A: If $F(a,b,c,d) = (w,x,y,z)$, note that $wz=xy$.  On the other hand, if $wz=xy$ and $z \ne 0$ then $a=x, b=z, c=y/z, d=1$ works.
