Notion of an inverse image as a function I've seen several variations of the inverse image of a function $f: A \to B$. Some define it in terms of elements, otherwise purely in terms of set. For example, if $C \subset B$, and $c \in C$, the definition may be given as
$$f^{-1} (c) = \{a \in A \mid f(a) = c\}$$
or as
$$f^{-1} \{c\} = \{a \in A \mid f(a) = c\}.$$
I have also seen this as one of the justifications for why the inverse image is not a function. But suppose that we restricted ourselves to only the second of these interpretations. In that case, $f$ mapped from subsets of $B$ to subsets of $A$. (That is, a subset is taken as a single object, so ouputting a set with multiple objects in the case where $f$ is not injective fine.) In the case where every element of some subset $C \subset B$ is not in the image of $f$, we define the inverse image to be the empty set.
My question is, do these restrictions turn the inverse image into a well-defined function? Is there a problem with defining this function as a mapping from subsets to subsets? Given the above interpretation, I could even potentially define $f^{-1}$ as a mapping from power set to power set.
 A: That's a perfectly well defined function from $B \to \mathscr P(A)$ but it is not a function from $B\to A$ at all.  So that defeats the purpose.
If we try to argue that a dolphin is a fish and then say, well, a dolphin is a mammal that lives the entirety of its life in water, so we can use that instead, is to .... be totally irrelevant.
You seem to think $g{c}$ vs. $g(c)$ vs $g(\{c\})$ is significant in whether we are mapping a set with a single element or single value.  And I suppose conceptually and in strict definition it is. But  in praticality, there isn't much difference (and in ZFC where everything is a set, there is no difference) between expressing somthing in terms of a single value or in terms of a set containing just the single value.  The actual notation $f(x)=c$ is shorthand for $(x, c)\in f \subset A\times B$.
Although $f^{-1}: B \to \mathscr P(A)$ is a well defined function it does not map subsets of $B$ to subsets of $A$.  It maps single elements (or equivalently singleton sets) of $B$ to subsets of $A$.  To map subsets of $B$ to subset of $A$ we must define $f^{-1}(C)$ where $c \subset B$ as $f^{-1}(C)$ as $\cup_{c\in C} \{a\in A| f(a) = c\}$.
With that definition $f^{-1}: \mathscr P(B) \to \mathscr P(A)$
A: No. Consider $A=\{1,2\}$ and $B=\{3,4\}$. with $f(a)=3$, a constant function. The inverse image restricted to $f^{-1}(3)=\{1,2\}$ and so it is not a well-defined function. You need the function to be injective if you want a elementwise inverse to exist.
