set functions and relations This might be a very basic question, but I hope you can help.
Given a (total) function $f\colon \wp(X) \rightarrow \wp(Y)$ under what condition does there exists a relation $g \subseteq X \times Y$ such that the image of $g[A] = f(A)$ whenever $A \in \wp(X)$? 
Is it enough to request distributivity, i.e., $f(A \cup B) =f(A) \cup f(B)$ whenever $A,B\in \wp(X)$ ? or do I need something more/else?
 A: Union preservation guarantees only that $f^{-1}(A\cup B)\subset f^{-1}(A)\cup f^{-1}(B)$ for any $A, B\in \wp(Y)$ while it is $g^{-1}[A\cup B]=g^{-1}[A]\cup g^{-1}[B]$ 
So it must be imposed the inverse preservation of the union as well.
$$f^{-1}(A\cup B)=f^{-1}(A)\cup f^{-1}(B)$$
A: What you give is not sufficient. Consider the mapping $f: \mathcal P(\mathbb N) \to \mathcal P(\{*\})$, such that
$$
f(A) = \begin{cases}
\mathbb \{*\} &\text{if $A$ is infinite}\\
\emptyset & \text{else.}
\end{cases}
$$
Clearly, $f(A \cup B) = f(A) \cup f(B)$ for all $A, B \subseteq \mathbb N$: if both $A,B$ are finite, then both sides of the equation are $\emptyset$, while if at least one is infinite, then both sides of the equation are $\{*\}$. However, there is no relation $g$, for if $g = \emptyset$, then trivially $g[\mathbb N] = \emptyset \neq \{*\} = f(\mathbb N)$; while if $g \neq \emptyset$, then $(x,*) \in g$, so that $g[\{x\}] = \{*\} \neq \emptyset = f(\{x\})$.
A sufficient condition on $f$ would be that $f$ distributes over all unions, so that
$$
f\left(\bigcup_{i \in I} A_i\right) = \bigcup_{i \in I} f(A_i)
$$
for all $\{A_i \mid i \in I\}$. In that case, we can define
$$
g = \{(x,y) \in X \times Y \mid y \in f(\{x\})\}.
$$
Then for each $A$, we have that
$$
f(A) = f\left( \bigcup_{x \in A} \{x\}\right) = \bigcup_{x \in X} f(\{x\}) = \bigcup_{x\in X}g[\{x\}] = g[X].
$$
Note that this is a very strong condition on $f$, and perhaps a weaker nice condition might be available.
A: Necessary conditions must be (for any $A,A_t\in\wp(X)$ and $B,B_t\in\wp(Y)$, where $t, t_1, t_2\in T$, $T$ an index set):
$$f(\cup_t A_t)=\cup_t f(A_t)\tag{1}$$
$$f^{-1}(\cup_t B_t)=\cup_t f^{-1}(B_t)\tag{2}$$
$$f(\cap_t A_t)\subseteq\cap_t f(A_t)\tag{3}$$
$$f^{-1}(\cap_t B_t)\subseteq\cap_t f^{-1}(B_t)\tag{4}$$
$$f(A_{t_1}\setminus A_{t_2})\subseteq f(A_{t_1})\setminus f(A_{t_2}\tag{3a})$$
$$f^{-1}(B_{t_1}\setminus B_{t_2})\subseteq f^{-1}(B_{t_1})\setminus f^{-1}(B_{t_2}\tag{4a})$$
$$A\subseteq f^{-1}(f(A))\tag{5}$$
$$B\subseteq f(f^{-1}(B))\tag{6}$$
($(3a)$ and $(4a)$ are consequence of $(3)$ and $(4)$ respectively)
Sufficiency of $(1)$, $(2)$, $(3)$, $(4)$, $(5)$, and $(6)$ must be proved.
