Why a function whose domain is a proper class does not have a codomain? On the Wikipedia article for codomain, in the third paragraph, it roughly says:

When the domain of a function is a
proper class X, in which case there is
formally no such thing as a triple (X,
Y, F). (?)  With such a definition
functions do not have a codomain.

As a proper class is a class that cannot be a member of some class, i.e. cannot be a set, I was wondering why a function with its domain being a proper class does not have a codomain?
Thanks and regards!
 A: My guess is: proper classes cannot belong to sets, and triples are sets, so the triple in question does not make sense. 
A: Actually, if you can define proper classes as such, you are probably in a theory with classes like NBG, so you have the "class comprehension" axiom schema which says that any formula that does not quantify over classes actually defines a class.
Let $F$ be a function defined on a proper class $X$, and let $\phi$ the formula (with free variable $y$ and quantified variable $p$)
$$
\exists p:p\in F \wedge \pi_2(p)=y
$$
where $\pi_2$ is the projection on the second component:
$$
\pi_2(p) = w \equiv \exists t\in p:(w\in t)\wedge (\forall (r,s \in p)\; r\neq s \implies w\notin r \vee w\notin s)
$$
or as finite set operations:
$$
⋃\left( ⋃p \setminus ⋂p \right)
$$
This formula defines a new class, that is in fact the codomain of $F$.
In general, classes cannot be put into sets, but you only need sets when there is an arbitrary number of classes, for instance you can define the product of two classes as a formula
$$
X\times Y = \{z: \pi_1(z)\in X \wedge \pi_2(z)\in Y\}
$$
where $\pi_1$ and $\pi_2$ are the usual projections.
