I asked a shorter version of this question at [math.stackexchange.com](https://math.stackexchange.com/questions/532560/inverses-of-two-argument-functions-with-respect-to-one-argument) 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)


1. 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.
2. $f\_1^{-1}$ is increasing in its first argument $\iff$ $f$ is increasing in its first argument.
3. $f\_2^{-1}$ is increasing in its second argument $\iff$ $f$ is increasing in its second argument.
4. $f\_1^{-1}$ 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?



Inverses of two-argument functions with respect to one argument
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.




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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](http://en.wikipedia.org/wiki/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.


