I need some help with this problem, if sombody could give me any idea of how to solve it (not the solution itself, but it would be better) I will appreciate it:
for a function $f: A → B$, prove $ ∀ Z ⊂ B$ that $ f(f^{-1}(Z)) ⊂ Z$ and that $f$ is surjective iff $ f(f^{-1}(Z)) = Z$
I already did the first one, but it may be important for the second one, wich is the one I have problem with. I know that the idea of proving surjectiveness is, for an arbitrary $y$ in the codomain $B$ of $f$, to look for an $x \in A$ that satisfies $f(x) = y$, set, but I don´t have a matching rule for $f: A \to B$, so I am kind of lost, any help is welcome, sorry for any mistakes in the grammatical part and the code knowledge.
$f(X)$: image in $B$ of a subset $X \subset A$
$f^{-1}(Y)$: inverse image in $A$ of the subset $Y \subset B$
Other tool I have: $ w ∈ f^{-1}(B) \Leftarrow \Rightarrow f(w) ∈ B $
our usual language and notation for functions and sets involves certain short-cuts which are often referred to as "abuses which should cause no confusion". this is permissible since brevity in notation can be a great aid to thinking. however sometimes there is a little confusion, and to make this less likely it is worth going through an occasional exercise in using a little more rigor than is normally required.
given two sets, $A$ and $B$, which we assume non-empty, we know how to form the set of ordered pairs $$A \times B =\{(a,b):a \in A, b \in B\}$$
a (general) relation from $A$ to $B$ is a subset of $A \times B$.
a function $f:A \to B$ is a special kind of relation in which each element of $A$ occurs exactly once. this simple restriction, which creates an asymmetry between the roles of $A$ and $B$, is remarkable effective, as the history of mathematics amply demonstrates. like the air we breathe, we are so used to having it around that we often hardly notice its existence.
to say $f$ is a function $f:A \to B$ demands very little of $B$, except that (unless $A=\varnothing$) at least one element of $B$ occurs as the second component of a pair $(x,b) \in f$.
the notion of a set $A$ is intimately bound up with the existence of a derived set, whose elements are the subsets of $A$ (including the empty set). this derived set, the power set of, $A$ is usually written $\mathfrak{P}(A)$, and its existence is one of the axioms of standard set theory.
now a function $f:A \to B$ gives rise in a natural way to two further functions: $$ f_*:\mathfrak{P}(A) \to \mathfrak{P}(B) $$ and $$ f^*:\mathfrak{P}(B) \to \mathfrak{P}(A) $$ the action of these related functions is defined by: $$ \forall X \subset A, f_*(X) = \bigcup_{x \in X} \{f(x)\} \subset B\tag{1} $$ and $$ \forall Y \subset B, f^*(Y) = \bigcup_{x \in A: f(x) \in Y}\{x\} \subset A \tag{2} $$
normally $f_*$ is simply written with the same symbol $f$ which is the base function the power set function $f_*$ is derived from. likewise $f^*$ is written as $f^{-1}$, a familiar and convenient, but also potentially confusing notation, since it is used to denote various quite different concepts in various fields of mathematics - all in some way concerned with inverses, which are often the place where certain difficulties become apparent.
if you are new to the game, it is worth spending a little time getting used to the implications of the definitions (1) and (2), simple as they are in essence. in particular, notice that in both cases we have formed a union over singleton sets. these definitions by means of unions over singletons actually help to make the logic clear, and provide the basic tools you need for answering questions like the ones you have asked here.
again it is common practice to make no notational distinction between an element $x$ and the singleton set $\{x\}$. this streamlines thought and writing, but the underlying distinction is also quite fundamental. it may often be ignored, but should never be entirely lost sight of.
the distinction matters if we look at the implication that $f_*$ is a function. it must therefore map every element of its domain $\mathfrak{P}(A)$ and this domain includes the empty set. so whereas the expression $f(\varnothing)$ is nonsensical, we can see using definition (1) that $f_*(\varnothing)=\varnothing$, and using definition (2) we have $f^*(\varnothing)=\varnothing$