Let $f : K \rightarrow L$ be a map of sets. Let $f^* : \mathscr{P}(L) \rightarrow \mathscr{P}(K)$ be the inverse image functor, sending a subset $B$ to its preimage under $f$. Consider $\mathscr{P}(L)$ and $\mathscr{P}(K)$ as ordered sets (via inclusion) -- then $f^*$ is order preserving. Then, further considering $\mathscr{P}(L)$ and $\mathscr{P}(K)$ as categories makes $f^*$ into a functor.

A left adjoint to $f^*$ is the the direct image functor, $f_* : \mathscr{P}(K) \rightarrow \mathscr{P}(L)$, sending a subset $A \subseteq K$ to its image under $f$. We claim $f_* \dashv f^*$. The unit $\eta$ is the statement "for all $A \in \mathscr{P}(K)$, $A \subseteq f^* f_* A$", which is clearly true. Likewise, the counit is the statement "for all $B \in \mathscr{P}(L)$, $f_* f^* B \subseteq B$", clearly true as well.

The right adjoint is much harder. We will denote it $f_*^c$ and call it the "complemented image functor". Define $f_*^c(A) = f_*(A^c)^c$, where $(\cdot)^c$ denotes the complement in the appropriate set ($K$ or $L$). The unit and counit statements are a bit nontrivial and require proof. The unit says "For all $B \in \mathscr{P}(L)$, $B \subseteq f_*^c f^* B$". This is non-trivial, and we show its truth via the following chain of implications:

$$\begin{align*} & b \not\in f_*^c f^* B \\ \Rightarrow \enspace & b \not\in f_*(f^*(B)^c)^c \\ \Rightarrow \enspace & b \in f_*(f^*(B)^c) \\ \Rightarrow \enspace & \exists x \in f^*(B)^c \tst f(x) = b \\ \Rightarrow \enspace & \exists x \not\in f^*(B) \tst f(x) = b \\ \Rightarrow \enspace & \exists x \tst f(x) \not\in B \tand f(x) = b \\ \Rightarrow \enspace & b \not\in B \end{align*}$$

So by contrapositive we have the unit. the counit is the statement "For all $A \in \mathscr{P}(K)$, $f^* f_*^c A \subseteq A$". We show this directly:

$$\begin{align*} & a \in f^* f_*^c A \\ \Rightarrow \enspace & a \in f^*(f_*(A^c)^c) \\ \Rightarrow \enspace & f(a) \in f_*(A^c)^c \\ \Rightarrow \enspace & f(a) \not\in f_*(A^c) \\ \Rightarrow \enspace & \forall x \in A^c, f(x) \neq f(a) \\ \Rightarrow \enspace & \forall x, x \not\in A \Rightarrow f(x) \neq f(a) \\ \Rightarrow \enspace & \forall x, f(x) = f(a) \Rightarrow x \in A \\ \Rightarrow \enspace & a \in A \end{align*}$$

thus, the unit and counit statements are true, and we have an adjunction $f^* \dashv f_*^c$.

So, $f^*$ has left and right adjoints: $f_* \dashv f^* \dashv f_*^c$.

Now, let's apply this to a particular situation. Let $X$ and $Y$ be sets and $p : X \times Y \rightarrow X$ be projection. We will regard a subset $S \subseteq X$ as a predicate $S(x)$ that is true on the members of $S$. Similarly subsets $R \subseteq X \times Y$ are 2-predicates $R(x,y)$.

We wish to determine what the left and right adjoints from the previous construction, $p_*$ and $p_*^c$, represent in terms of predicates. This is really just a matter of interpreting the definitions in the language of logic. Given a 2-predicate $R(x,y)$, one can see that $p_* R$ is the predicate

$$\exists y \in Y \tst R(x,y)$$

which is a predicate in one variable $x \in X$. Since $p_*^c$ is defined in terms of $p_*$ in the first place, we can simply plug it in and interpret complement as "not" ($\neg$). So $p_*^c R$ is the predicate

$$\neg (\exists y \in Y \tst \neg R(x,y))$$

Of course, this reduces to

$$\forall y \in Y, R(x,y)$$

So, when we take $p$ as the projection onto $X$, $p_*$ corresponds to the operation of placing "$\exists y \in Y$" in front of the input, and $p_*^c$ places "$\forall y \in Y$" in front of the input.