Let $f'\colon X \to Y$ and $P(X)$ and $P(Y)$ be the powersets of $X$ and $Y$.
Let $f \colon P(X) \to P(Y)$,
$f(U) = \{ y \in Y | \exists x \in U (f'(x) = y) \},$
and $f^{-1} \colon P(Y) \to P(X)$,
$f^{-1}(V) = \{ x \in X | \exists y \in V (f'(x) = y) \}.$
Then $(f,f^{-1})$ forms a Galois connection between $P(X)$ and $P(Y)$, when both of these sets are ordered by inclusion.
How to prove (simply) that $f(U) \subseteq V$ if and only if $U \subseteq f^{-1}(V)$?
If we assume that $f(U) \subseteq V$, then $f(U)$ includes elements $y \in Y$ so that $f'(x)=y$ for some $u \in U$ and $V$ includes all of these elements and maybe more. But the set $f^{-1}(V)$ includes only the elements $x \in X$ which for $f'(x)=y$ for some $y \in V$. So obviously $U \subseteq f^{-1}(V)$. But this is kind of a mess, how to prove this in a more simple way?
Let's assume that $f(U) \subseteq V$. Then $f^{-1}(f(U)) \subseteq f^{-1}(V)$. It is known that $U \subseteq f^{-1}(f(U))$, so now $U \subseteq f^{-1}(V)$, because the relation $\subseteq$ is transitive. Hence, if $f(U) \subseteq V$ then $U \subseteq f^{-1}(V)$.
And same works the other way around, assuming that $f(f^{-1}(V)) \subseteq V$ is known.