TITLE: Inverses of two-argument functions with respect to one argument
QUESTION [1 upvotes]: I asked a shorter version of this question at math.stackexchange.com four days ago but it hasn't gotten any answers or comments.
Consider a function $f : A \times B \to C$ and two inverses, each with respect to one argument; i.e. $f_1^{-1}$ and $f_2^{-1}$ defined such that $f(x,y)=z \iff f_1^{-1}(z,y)=x \iff f_2^{-1}(x,z)=y$. A simple example is addition:
$$
\begin{aligned}
f(x,y) &= x+y \\
f_1^{-1}(z,y) &= z-y \\
f_2^{-1}(x,z) &= z-x
\end{aligned}
$$
Such collections can of course be generalized to any number of arguments.
They have some interesting properties; e.g. if $A$, $B$ and $C$ are totally ordered, then (I think)

If $f$ is surjective and strictly increasing or decreasing in each argument (for every value of the other argument), then $f_1^{-1}$ and $f_2^{-1}$ exist.
$f_1^{-1}$ is increasing in its first argument $\iff$ $f$ is increasing in its first argument.
$f_2^{-1}$ is increasing in its second argument $\iff$ $f$ is increasing in its second argument.
$f_1^{-1}$ is increasing in its second argument $\iff$ $f_2^{-1}$ is increasing in its first argument $\iff$ $f$ is increasing in its first or second argument, but not both.

Is there a name for $f_1^{-1}$ and $f_2^{-1}$? (It's not "partial inverse," though that would be a great name for them, analogous to "partial derivative.")
In general, what should I search for to find properties about such collections?

REPLY [1 votes]: Surjective and monotonic implies surjective and injective (i.e. bijective) but not vice versa. For example if $A=B=C$ is finite (and well ordered) then there is a unique  surjective increasing map (the identity) and a unique decreasing one.  So (for size more than two) there would not be enough maps to have them all be monotonic. However just requiring surjective and injective one essentially has a latin square. I suppose one could have a circular order (as in addition $\mod n$) although that would only give circulant latin squares (I think) from the rich universe of latin squares.

Some details are ambiguous and the notation could be improved (in my opinion). Is $f_1^{-1}$ intended to be defined on all of $C \times  B$ or merely on certain pairs? (And similarly for $f_2^{-1}.$) A useful example for clarification is exponentiation $f:A \times B \to C$ (with $A,B,C$ to be determined) when we would hope to have 
$$
\begin{aligned}
z=f(x,y) &= x^y \\
g(z,y)=f_1^{-1}(z,y) &= \sqrt[y]{z} \\
 h(x,z)=f_2^{-1}(x,z) &= \log_x(z)
\end{aligned}
$$ 
Suppose $A=C$ is the set of positive odd integers other than $1$ and $B$ is the set of all positive integers. Then the domain of $f_1^{-1}$ is all pairs $(z,y)$ where $z$ is an odd $y$th power. Would that suit you or should we instead try $A=C=(0,1) \cup (1,\infty)$  while $B=(-\infty,0) \cup (0,\infty)?$ (in which case better to say $g(z,y)=z^{1/y}$). 
Note: Applying logarithms to $A$ and $C$ this structure becomes multiplication of non-zero reals, however the exponential version seems clearer for telling the ingredients apart as in the next comments..

Using the example above we see that 
$g_1^{-1}(x,y)=f(x,y) $ and $g_2^{-1}(x,z)=h(x,z) $
$h_1^{-1}(y,z)=g(z,y) $ and $h_2^{-1}(x,y)=f(x,y) $
So the inverse notation seems more confusing than helpful. Perhaps it would be clearer to say (in the case that the intended domains are all of $C \times B$ and $A \times C$) that one has a set of triples $F \subset A \times B \times C$ such that the first two uniquely determine the third (So $F=\lbrace(x,y,f(x,y))\mid (x,y) \in A \times B \rbrace$) and wonder if any two determine the other. If (and only if) $f(a,\cdot): B \to C$ and $f(\cdot,b): A \to C$ are bijections (for each fixed value of $a$ or $b$) we have that any two entries determine the third. Since a monotonic map is injective, if the two restrictions are surjective and monotonic they are bijections. But a weaker condition is that they are surjective and injective.
I suggest the notation
$z=F_3(x,y)$ and $y=F_2(x,z)$ and $x=F_1(y,z)$ 
or else  
$z=F(x,y,-)$ $y=F(x,-,z)$ and $x=F(-,y,z)$

As a final comment, If the sets are finite then they have the same size, could be taken as all being $[n]=\lbrace 1,2,\cdots,n\rbrace$ and one has a square with $(i,j)$ entry $f(i,j).$ If each row and column is a permutation of $[n]$ then it is a latin square.  I think that the generalization you want to a larger number of arguments would be (in the finite case) a system of mutually orthogonal latin squares.
{"set_name": "stack_exchange", "score": 1, "question_id": 145676}