Functor between ordered sets. 
(a) Let $f : K \rightarrow L$ be a map of sets, and denote by $f^* : \mathscr{P}(L) \rightarrow \mathscr{P}(K)$ the map sending a subset $S$ of $L$ to its inverse image $f^{-1 }[S] \subseteq K$. Then $f$  is order-preserving with respect to the inclusion orderings on $\mathscr{P}(K)$ and $\mathscr{P}(L)$, and so can be seen as a functor. Find left and right adjoints to $f$ .
(b) Now let $X$ and $Y$ be sets, and write $p: X\times Y \rightarrow X$ for first projection. Regard a subset $S$ of $X$ as a predicate $S(x)$ in one variable $x\in X$, and similarly a subset $R$ of $X\times Y$ as a predicate $R(x,y)$ in two variables. What, in terms of predicates, are the left and right adjoints to p? For each of the adjunctions, interpret the unit and counit as logical implications. (Hint: the left adjoint to $p^*$ is often written as  , $\exists_Y$ and the right adjoint as $\forall_Y$.)

I solved (a): i find that defining $f^{**},f^{+}:\mathscr{P}(K)\rightarrow\mathscr{P}(L) $ as $f^{**}(A)=f[A]$ and $f^{+}(A)=L-f[K-A] $ we get $f^{**}\dashv f^*\dashv f^+$. The equations
$$A\subset f^* f^{**}(A)$$
$$f^{**}f^*(B)\subset B$$
and
$$B\subset f^+ f^{*}(B)$$
$$f^*f^+(A) \subset A$$
prove the claim above since these equations are the unit and counit in each adjunction.
I have some questions related with (b). Firt at all I have to understand the adjoints as predicates
$$p^{**}(R)(x)=\begin{cases} 1 & \text{ si }& \exists y\in Y\, (R(x,y))\\0 & \text{otherwise}& \end{cases}$$
$$p^+(R)(x)=\begin{cases} 1 & \text{ si }& \forall y\in Y\, (R(x,y))\\0 & \text{otherwise}& \end{cases} $$
Then I think that we can interprete $p^{**}p^*(B)\subset B$ as the logical implication $\exists y\in Y\, (B\times Y)(x,y)\rightarrow B(x)$ and $B\subset p^+ p^*(B)$ as the logical implication $B(x)\rightarrow \forall y\in Y (A\times Y)(x,y)$. Are these results correct?
Otherwise, I do not know a suitable logic interpretation for $A\subset p^*p^{**}(A)$ and $p^*p^+(A)\subset A$. Anyone knows?
Thanks
 A: The first thing you could do to try to verify the results is to check if your interpretations produce true formulas.  So the first question is: Are
$$(\exists y \in Y. (B\times Y)(x,y)) \to B(x) \\
  A(x) \to \forall y \in Y. (A\times Y)(x,y)$$
true as logical formulas?  (Note, be careful and generous with your parentheses when talking to people who might not share your notational conventions.)
Of course, even if the answer to that is "yes", that's not really a proof.  A proof would show that your counit/unit actually satisfies the laws of the counit/unit of an adjunction.  One way to do this would be to work out the unit/counit and verify the triangle identities.  Instead, I'm going to use the isomorphism of homsets view of adjunction.  That works out to essentially mutual entailment of two logical formulas for each adjunction.  For $\exists_Y$,
$$(\exists y \in Y.Q(x,y)) \to P(x) \iff Q(x,y) \to P(x)$$
and for $\forall_Y$,
$$P(x) \to Q(x,y) \iff P(x) \to \forall y \in Y. Q(x,y)$$
where this notation implicitly lifts a unary predicate to a binary predicate, so $P(x)$ is really $p^*(P)(x,y)$.  Actually, it may (or may not) look more familiar written like:
$$\frac{\exists y \in Y.Q(x,y) \vdash P(x)}{Q(x,y) \vdash P(x)} \quad\quad
\frac{P(x) \vdash Q(x,y)}{P(x) \vdash \forall y \in Y. Q(x,y)}$$
When I said I was going to prove this, I lied.  The above are practically part of the defining rules of existential and universal quantification.  In this case, the usual side condition "$y$ not free in $P$" is handled... well, in our case by being explicit about parameterization, but in either case by $p^*$.  $p^*$ is what adds $``y"$ to the context.  If that's not clear now, it will probably become clearer further in your study.  The counit/unit is when we consider the case induced by $P(x) \vdash P(x)$ in the context with $y$, i.e. $p^*(P)(x,y) \vdash p^*(P)(x,y)$, which gives:
$$(\exists y \in Y.P(x)) \to P(x) \\
P(x) \to \forall y \in Y.P(x)$$
