Power sets as a function Is the correspondence \begin{align*}f: \mathcal{P}(X)&\to \mathcal{P}(\mathcal{P}(X))\\A&\mapsto \mathcal{P}(A)\end{align*} a function?

If it is, I am thinking of using it to solve this problem

Let $(A_i)_{i\in I}$ be a family of subsets of a set $X$. Are the relations:
  $$\mathcal{P}\big(\bigcap_{i\in I}A_i\big) = \bigcap_{i\in I}\mathcal{P}(A_i)\\\mathcal{P}\big(\bigcup_{i\in I}A_i\big) = \bigcup_{i\in I}\mathcal{P}(A_i)$$ true?

If $f$ is a function, then the answer is quick and easy (it would need to be injective for the first relation to be true, and it seems it would be).
 A: Yes, it's a function. I don't see how it is useful for solving your problem. If I were faced with those questions, instead of trying to find some way to state them in more arcane and confusing language, I'd be trying to state them more simply and clearly. In plain language, the questions are:

Is it true that something is a subset of the intersection of a bunch of sets if and only if it's a subset of each of them?

  Is it true that a set is a subset of the union of a bunch of sets if and only if it's a subset of one of them?

Offhand I'd guess that one and a half of those statements are true.
P.S. Although $\mathcal P$ is a function, your idea for answering the questions is wrong. First let me explain your notational confusion.
Consider a function $f:X\to Y.$ If $a$ is an element of $X,$ the symbol $f(a)$ denotes (as usual) the value of the function at $a,$ and is an element of $Y.$ On the other hand, if $A$ is a subset of $X,$ then $f[A]$ (note the square brackets) denotes the set $\{f(a):a\in A\}$ and is a subset of $Y.$ 
We often ignore this distinction and use round brackets $f(\cdot)$ in both cases, and usually this does no harm. However, it gets confusing if the elements of $X$ are themselves sets, as they are here. In particular, the identity $f[A\cup B]=f[A]\cup f[B]$ (note square brackets) is always true, but $f(A\cup B)=f(A)\cup f(B)$ is usually false.
