Let $S$ and $T$ be partially ordered sets, where the order relations are both denoted with $\le$. If $\varphi\colon S\to T$ and $\psi\colon T\to S$ are order preserving maps such that $s\le\psi(\varphi(s))$ and $\varphi(\psi(t))\le t$, for all $s\in S$ and $t\in T$, then
$$
\varphi(s)=\varphi(\psi(\varphi(s))),\quad
\psi(t)=\psi(\varphi(\psi(t)))
$$
for all $s\in S$ and $t\in T$.
Proof: obvious. ;-) Well, from $s\le\psi(\varphi(s))$ we deduce $\varphi(s)\le\varphi(\psi(\varphi(s)))$ because $\varphi$ is order preserving. On the other hand, setting $t=\varphi(s)$, we know that $\varphi(\psi(t))\le t$, which is the same as saying that $\varphi(\psi(\varphi(s)))\le\varphi(s)$.
The verification for the other equality is just the same.
Now, let $S=P(A)$ and $T=P(B)$ be the power sets, with the order relation on them being the inclusion. We set $\varphi(X)=f[X]$ and $\psi(Y)=f^{-1}[X]$, for $X\in P(A)$ and $Y\in P(B)$. These maps are order preserving and
$$
X\subseteq\psi(\varphi(X))=f^{-1}[f[X]]
$$
while
$$
\varphi(\psi(Y))=f[f^{-1}[Y]]\subseteq Y
$$
Note that idea can be vastly generalized: if $F$ is a left adjoint of the functor $G$, then $FGF$ and $GFG$ are naturally equivalent to $F$ and $G$ respectively.
In the setting above, considering $S$ and $T$ as categories in the usual way, $\varphi$ and $\psi$ are functors because they are order preserving; the inequalities $s\le\psi(\varphi(s))$ and $\varphi(\psi(t))\le t$ are just the statement that $\varphi$ is a left adjoint of $\psi$.
The ordered set case is an example of a (monotone) Galois connection; in fact from the inequalities we can easily prove that, for $s\in S$ and $t\in T$, we have
$$
\varphi(s)\le t\quad\text{if and only if}\quad s\le \psi(t)
$$
Indeed, suppose $\varphi(s)\le t$; then $\psi(\varphi(s))\le\psi(t)$, so $s\le\psi(t)$. The converse is similar.
