Let $\mathcal P \left({A}\right)$ be the power set of $A$. Define a mapping $E: \mathcal P \left({A}\right) \to \mathcal P \left({A}\right)$ thus: : $E \left({S}\right) = A \setminus g \left({B \setminus f \left({S}\right)}\right)$ === $E$ is increasing === Let $S, T \in \mathcal P \left({A}\right)$ such that $S \subseteq T$. Then: {{begin-eqn}} {{eqn | l = f \left({S}\right)       | o = \subseteq       | r = f \left({T}\right)       | c = Image of Subset is Subset of Image }} {{eqn | ll= \implies       | l = B \setminus f \left({T}\right)       | o = \subseteq       | r = B \setminus f \left({S}\right)       | c = Set Difference with Subset is Superset of Set Difference }} {{eqn | ll= \implies       | l = g \left({B \setminus f \left({T}\right)}\right)       | o = \subseteq       | r = g \left({B \setminus f \left({S}\right)}\right)       | c = Image of Subset is Subset of Image }} {{eqn | ll= \implies       | l = A \setminus g \left({B \setminus f \left({S}\right)}\right)       | o = \subseteq       | r = A \setminus g \left({B \setminus f \left({T}\right)}\right)       | c = Set Difference with Subset is Superset of Set Difference }} {{end-eqn}} That is, $E \left({S}\right) \subseteq E \left({T}\right)$. {{qed|lemma}} By the Knaster-Tarski Lemma, $E$ has a fixed point $X$. By the definition of fixed point: : $E \left({X}\right) = X$ Thus by the definition of $E$: : $A \setminus g \left({B \setminus f \left({X}\right)}\right) = X$ Therefore: : $(1): \quad A \setminus \left({A \setminus g \left({B \setminus f \left({X}\right)}\right)}\right) = A \setminus X$ Since $g$ is a mapping into $A$: : $g \left({B \setminus f \left({X}\right)}\right) \subseteq A$ Thus by Relative Complement of Relative Complement: : $A \setminus \left({A \setminus g \left({B \setminus f \left({X}\right)}\right)}\right) = g \left({B \setminus f \left({X}\right)}\right)$ Thus by $(1)$: :$g \left({B \setminus f \left({X}\right)}\right) = A \setminus X$ Let $f' = f \restriction_{X \times f \left({X}\right)}$ be the restriction of $f$ to $X \times f \left({X}\right)$. Similarly, let $g' = g \restriction_{\left({B \setminus f \left({X}\right)}\right) \times \left({A \setminus X}\right)} = g \restriction_{\left({B \setminus f \left({X}\right)}\right) \times g \left({B \setminus f \left({X}\right)}\right)}$. By Injection to Image is Bijection, $f'$ and $g'$ are both bijections. Define a relation $h: A \to B$ by $h = f' \cup {g'}^{-1}$. We will show that $h$ is a bijection from $A$ onto $B$. The domain of $f'$ is $X$, which is disjoint from the codomain, $A \setminus X$, of $g'$. The domain of $g'$ is $B \setminus f \left({X}\right)$, which is disjoint from the codomain, $f \left({X}\right)$, of $f'$. Let $h = f' \cup {g'}^{-1}$. By the corollary to Union of Bijections with Disjoint Domains and Codomains is Bijection: : $h$ is a bijection from $X \cup \left({A \setminus X}\right)$ onto $f \left({X}\right) \cup \left({B \setminus f \left({X}\right)}\right)$. By Union with Relative Complement, $h$ is a bijection from $A$ onto $B$. Since $f' \subseteq f$ and $g' \subseteq g$, each element of $h$ is an element of $f$ or of $g^{-1}$. That is, if $y = h \left({x}\right)$ then either $y = f \left({x}\right)$ or $x = g \left({y}\right)$. {{Qed}}