Hello, 


I would like to know how to currify a partial mapping $A \times B \to\_p C$?


Here, by "partial mapping from $X$ to $Y$", I mean a function from a subset of $X$ to $Y$, as explained here
[Partial Function](http://en.wikipedia.org/wiki/Partial_function).


It is known that $(A \times B) \to C$ is isomorphic to $A \to B \to C$. The isomorphism is known as "currying". 


My question is about the relation between $A \times B \to\_p C$ and $A \to\_p B \to\_p C$, where ${\to\_p}$ denotes a partial function. Apparently, they are not isomorphic. But I would like to know how we can currify $A \times B \to\_p C$?


Thanks.


Edit: My question was really, can we say that $A \to\_p (B \to C) )$ is isomorphic to $A \times B \to\_p C$? (I think no.) Which set would then be isomorphic to $(A \to\_p (B \to C))$ (except itself)?



Currying a " partial mapping A*B -->_p C 
For plain vanilla sets and functions, a partial function $f$ from $A$ to $B$ can be viewed as essentially the same thing as a total function $g: A \to B + 1$, where the codomain is the disjoint union of $B$ with a 1-element set. The idea is that whenever $f(a)$ is undefined, we define $g(a)$ to be the element of $1$. 


Thus, if $Par(A, B)$ denotes the set of partial functions, and $\hom(A, B)$ the set of total functions, we have 


$$Par(A \times B, C) \cong \hom(A \times B, C+1) \cong \hom(A, \hom(B, C+1)) \cong \hom(A, Par(B, C))$$ 


and that tells you how to currify (curryfy?). 


For toposes, the situation is just slightly more complicated, but it remains true that partial morphisms into an object $B$ are representable (not by $B+1$ in general, but by something else). But the same currification carries over. 


**Edit:** I was asked to respond to: what about currying (or schoenfinkelifying!) the other way, to $Par(A, \hom(B, C))$. It pretty clearly gives a different result, just by cardinality considerations on finite sets; the difference is pretty stark by considering $A = 1$, where we compare $Par(B, C)$ to $\hom(B, C) + 1$ -- obviously the first is larger if $B$ has two or more elements. Sorry I can't say much more right now. 


