Let $\rho$ is a function mapping every binary relation $f$ (on some set $U$) into a function which maps binary relations into binary relations by the formula
$$(\rho(f))(g) = f\circ g.$$
Is $\rho$:
(The order assumed is the set-theoretic inclusion of binary relations.)
If it has adjoints, what these adjoints are?
On
For various reasons (probably largely to do with issues of variance), it's a little tricky interpreting the question. There is a classical notion of Galois connection which involves a pair of contravariant poset maps
$$\phi: P(X) \to P(Y), \qquad \psi: P(Y) \to P(X)$$
such that $A \leq \psi(\phi(A))$ for all $A \in P(X)$ and $B \leq \phi(\psi(B))$ for all $B \in P(Y)$. There is also a more general notion of Galois connection between posets, not just power sets, and is basically the notion of "pair of adjoint functors between posets", in other words covariant functors between posets $\phi: P^{op} \to Q$ and $\psi: Q \to P^{op}$ such that $\psi \dashv \phi$.)
I'll go with the second meaning. I assume the set of functions from binary relations to binary relations $\hom(P(X \times X), P(X \times X))$ is given the "pointwise" order, where $f \leq g$ means $f(A) \leq g(A)$ in $P(X \times X)$ for all $A \in P(X \times X)$, taking $\leq$ to be the usual inclusion order on $P(X \times X)$.
Then the covariant map $\phi: P(X \times X) \to \hom(P(X \times X), P(X \times X))$ in the question seems to take a relation $R$ to the function $S \mapsto R \circ S$, also denoted $\lambda S. R \circ S$. Then, as I think porton is surmising, $\phi$ is indeed a left adjoint. One way of demonstrating that is via the adjoint functor theorem for functors between complete posets: that a functor in this case is a left adjoint if and only if it preserves arbitrary joins = unions.
To prove that $R \mapsto \lambda S. R \circ S$ preserves unions, we just check this pointwise, i.e., that for each $S$ the map $R \mapsto R \circ S$ preserves unions. And indeed it does. That in part is what porton is verifying in his answer.
My own preferred proof is direct. Observe the functor $- \circ S: P(X \times X) \to P(X \times X)$ has a right adjoint, namely the map $S \Rightarrow (-): P(X \times X) \to P(X \times X)$ defined by
$$P(X \times X) \ni T \mapsto \{(x, y): \forall_z \; y S z \Rightarrow x T z\}.$$
The adjunction just says
$$\exists_y \; x R y \; \wedge\; y S z \Rightarrow x T z\;\;\; \text{iff}\;\;\; x R y \Rightarrow \forall_z \; y S z \Rightarrow x T z$$
which is obviously the case. Now we calculate, for all $f \in \hom(P(X \times X), P(X \times X))$,
$$\lambda S. R \circ S \leq f \;\; \text{iff}\;\; \forall_S \; R \circ S \leq f(S) \;\;\text{iff}\;\; \forall_S \; R \leq S \Rightarrow f(S) \;\;\text{iff}\;\; R \leq \bigcap_{S: P(X \times X)} S \Rightarrow f(S);$$
this shows directly that $R \mapsto \lambda S. R \circ S$ is a left adjoint.
But it is not a right adjoint; indeed, it does not preserve arbitrary meets. This boils down to saying that there are $S$ such that $- \circ S$ does not preserve arbitrary meets, so porton's second proposition is false. It doesn't even preserve binary meets. It is true for $S$ atomic (so porton's lemma is okay); this is actually a nice fact, that composing with an atomic relation preserves arbitrary meets. But not for general $S$.
For a specific example, consider a set with four elements $u, v, v', w$. Let $R$ be the atomic relation $\{(u, v)\}$, let $R'$ be the atomic relation $\{(u, v')\}$, and let $S$ be the relation $\{(v, w), (v', w)\}$. Then $(R \wedge R') \circ S$ is the empty relation (there is no $(x, z)$ such that $\exists_y x R y \wedge x R' y \wedge y S z$), whereas we do have $u R \circ S w$ and $u R' \circ S w$. Hence the statement
$$(R \circ S) \wedge (R' \circ S) \leq (R \wedge R') \circ S$$
is false in this example. (I suspect the problem is that porton was misapplying the distributivity of meets over unions.)
Let $F$ is a set of binary relations and $A$ is a binary relation.
Proposition $\left( \bigcup F \right) \circ A = \bigcup \left\{ f \circ A \,|\, f \in F \right\}$.
Proof $( x ; z) \in \left( \bigcup F \right) \circ A \Leftrightarrow \exists y : \left( ( x ; y) \in A \wedge ( y ; z) \in \bigcup F \right) \Leftrightarrow \exists y : ( ( x ; y) \in A \wedge \exists f \in F : ( y ; z) \in f) \Leftrightarrow \exists y, f \in F : ( ( x ; y) \in A \wedge ( y ; z) \in f) \Leftrightarrow \exists f \in F \exists y : ( ( x ; y) \in A \wedge ( y ; z) \in f) \Leftrightarrow \exists f \in F : ( x ; z) \in f \circ A \Leftrightarrow ( x ; z) \in \bigcup \left\{ f \circ A \,|\, f \in F \right\}$.
Proposition $\left( \bigcap F \right) \circ a = \bigcap \left\{ f \circ a \,|\, f \in F \right\}$ if $a$ is a singleton binary relation.
Proof Let $a = \{ ( a_0 ; a_1) \}$.
$( x ; z) \in \left( \bigcap F \right) \circ a \Leftrightarrow x = a_0 \wedge ( a_1 ; z) \in \bigcap F \Leftrightarrow x = a_0 \wedge \forall f \in F : ( a_1 ; z) \in f \Leftrightarrow \forall f \in F : ( x = a_0 \wedge ( a_1 ; z) \in f) \Leftrightarrow \forall f \in F : ( x ; z) \in f \circ a \Leftrightarrow ( x ; z) \in \bigcap \left\{ f \circ a \,|\, f \in F \right\}$.
Proposition (wrong) $\left( \bigcap F \right) \circ A = \bigcap \left\{ f \circ A \,|\, f \in F \right\}$.
This proposition fails with the following counter-example. It seems that in the proof I have misused the law of complete distributivity.
$A = \{(4,2),(4,3)\}$; $f_1 = \{(2,1)\}$; $f_2 = \{(3,1)\}$; $F=\{f_1,f_2\}$.
Then $(\bigcap F) \circ A = \varnothing$ but $\bigcap \{f \circ A \,|\, f\in F\} = \{(4,1)\}$.
As a consequence from the above we have:
$\rho \bigcup F = \bigcup \rho [F]$, that is $\rho$ has upper adjoint.
From $\left( \bigcap F \right) \circ a = \bigcap \left\{ f \circ a \,|\, f \in F \right\}$ it follows that $\rho$ has a lower adjoint, because to specify $\rho$ it is enough to specify it on singleton relations.
The remaining problem is to write explicit formulas for lower and upper adjoint of $\rho$.
I don't specify our ``universal sets'' in the below formula for simplicity. You can easily correct this omission if you want to be formal.
The upper adjoint is $b \mapsto \max \left\{ x \, | \, \forall a : x \circ a \subseteq b [ a] \right\}$.
The lower adjoint is $b \mapsto \min \left\{ x \, | \, \forall a : x \circ a \supseteq b [ a] \right\}$.
Can these formulas be simplified?