Theorem. For surjectivity of a mapping $f\!:\,A \to B$ , it's necessary and sufficient that

(1)

Proof. $1^{\underline{o}}$ . Suppose that $f\!:\,A \to B$ is surjective. Let $X$ be an arbitrary subset of $A$ and $y$ any element of the set $B\!\smallsetminus\!f(X)$ . By the surjectivity, there is an $x$ in $A$ such that $f(x) = y$ , and since $y \notin f(X)$ , the element $x$ is not in $X$ , i.e. $x \in A\!\smallsetminus\!X$ and thus $y = f(x) \in f(A\!\smallsetminus\!X)$ . One can conclude that $B\!\smallsetminus\!f(X) \,\subseteq\, f(A\!\smallsetminus\!X)$ for all $X \subseteq A$ .

$2^{\underline{o}}$ . Conversely, suppose the condition (1). Let again $X$ be an arbitrary subset of $A$ and $y$ any element of $B$ . We have two possibilities:
a) $y \notin f(X)$ ; then $y \in B\!\smallsetminus\!f(X)$ , and by (1), $y \in f(A\!\smallsetminus\!X)$ . This means that there exists an element $x$ of $A\!\smallsetminus\!X \subseteq A$ such that $f(x) = y$ .
b) $y \in f(X)$ ; then there exists an $x \in X \subseteq A$ such that $f(x) = y$ .
The both cases show the surjectivity of $f$ .