What is the proper usage of $f: X \to Y$ and $f: \mathcal{P}(X) \to \mathcal{P}(Y)$ in proof writing.

Question

I have read somewhere that suppose we are given a $$f: X \to Y$$ then $f$ is further associated with $$f: \mathcal{P}(X) \to \mathcal{P}(Y)$$

Does "associated" here means extended to a set valued mapping? Why should $f$ always be associated with a set valued map?

But more importantly, why is it in proofs, I always see the usage of this latter set valued map without any warning.

For example, let $f: X \to Y$ be continuous, then $f^{-1}(U)$ is open, given $U$ is open $Y$, but $f(V)$ is not necessarily open for any open $V$ in $X$.

We almost always define what $f$ is, i.e. $f: X \to Y$. But never give any clarification or warning that we are changing the definition of $f$.

Answer 1

It's overloaded notation, for sure. But I have never seen it mean anything other than the following:

Let $f : X \to Y$. Then, for $A \subseteq X$, $f(A) := \{ f(x) : x \in A \}$.

The reason there isn't much notice is probably because it's a very common convention. It may even be explicit at some point in some books on Real Analysis.

It turns out that set theorists are usually more explicit in the difference by letting $f[A] = \{ f(x) : x \in A\}$ so one is not confused about the domain of the $f$ in question.

Answer 2

The definition of $f$ isn't changing. $f(U)$ is simply standard notation for the image of $U$ under $f$, and similarly for the inverse image $f^{-1}(U)$. Of course $f:X \to Y$ is not the same function as its direct image function, which you could denote $\mathcal{P}(f):\mathcal{P}(X) \to \mathcal{P}(Y)$ if you really wanted.

If category theory interests you, we can interpret the power set as an endofunctor of the category of sets. It associates to a set $X$ its power set $\mathcal{P}(X)$, and to a morphism (function) $f:X \to Y$ the direct image function $\mathcal{P}(f):\mathcal{P}(X) \to \mathcal{P}(Y)$. There is similary a contravariant power set functor which maps a function to its inverse image function.

Answer 3

Some authors use different notation, i.e., for $f\colon A\to B$, the associated direct image function is denoted $f_\to\colon \mathcal P(A)\to \mathcal P(B)$. However, since this association is very common it is standrd to simply write $f$ for the extended set valued function, without warning as you say.

Some care must be exercised, since for instance if $A=\{1, \{1\}\}$ and $f\colon A\to \mathbb N $ is given by $f(1)=17$ and $f(\{1\})=243$, then there is a big difference between $f_\to (\{1\})$ and $f(\{1\})$. Basically, when no notational distinction is made, we assume context determines the precise meaning.