Let $A$ and $B$ be arbitrary sets with arbitrary elements $a\in A$ and $b\in B$, and powersets $\mathcal{P}(A)$ and $\mathcal{P}(B)$. Further, let $f:A\to\mathcal{P}(B)$ be a map; and for any set $f(a)$, let $\mathcal{P}(f(a))$ be its powerset. Now,
- Define a map $g:A\to \mathcal{P}(B)$ that satisfies $g(a)\in\mathcal{P}(f(a))$ for all $a\in A$.
- Define a map $g’:\prod_{a\in A}a\to\prod_{a\in A}\mathcal{P}(f(a))$.
My question is: is it true that $g\equiv g’$? I am trying to figure out whether $g$ and $g’$ are equivalent definitions of the same mathematical object.
EXAMPLE [addition following comments]:
Consider the following example. Let $A\equiv\{a_1,a_2\}$ and $B\equiv\{b_1,b_2,b_3\}$. Further, let $f:A\to\mathcal{P}(B)$ to be such that $f(a_1)=\{b_1,b_3\}$ and $f(a_2)=\{b_2,b_3\}$.
Then, define $g:A\to\mathcal{P}(B)$ to be such that \begin{gather} g(a_1)=\{b_1,b_3\}\\ g(a_2)=\{b_2\} \end{gather}
Then, define $g’:\prod_{a\in A}a\to\prod_{a\in A}\mathcal{P}(f(a))$ to be such that \begin{gather} g’(\{(a_1,a_2)\})=(\{b_1,b_3\},b_2)\\ \end{gather}
Then, in a way, one can perhaps say that $g$ and $g’$ represent the same mathematical object, since for each $a\in A$, the restriction of $g’(\cdot)$ to $a$ coincides with $g(a)$ (i.e., $g(a)=g’_a(\cdot)$).
Thank you all.