Let $\Phi\subseteq (A\times A)\times(A\times A)$, $F_0,F_1\subseteq A\times A$ (for some set $A$) be binary relations.
I will denote $\pi_0$ and $\pi_1$ the projections of a cartesian product of sets.
Do inequalities $\pi_0\circ\Phi \subseteq F_0\circ\pi_0$ and $\pi_1\circ\Phi \subseteq F_1\circ\pi_1$ imply existence of a binary relation $\Psi\supseteq\Phi$ (we also require $\Psi\subseteq (A\times A)\times(A\times A)$) such that $\pi_0\circ\Psi = F_0\circ\pi_0$ and $\pi_1\circ\Psi = F_1\circ\pi_1$?
If yes, is this $\Psi$ unique?
Remark: For binary relations $f$ and $g$ I define the binary relation $g\circ f$ by the formula:
$(x,z)\in g\circ f \Leftrightarrow \exists y: ((x,y)\in f\wedge(y,z)\in g)$.
No, such a set does not always exist. Here's a counterexample: Let $A=2$, let $F_0=\{(0,1)\}$ and $F_1=\{(0,1)\}$ and let $\Phi=\varnothing$. Then $((1,0),1)\in F_1\circ\pi_1$. If there is such a $\Psi$ then $((1,0),1)\in\pi_1\circ\Psi$, i.e. there is some $n\in 2$ such that $((1,0),(n,1))\in\Psi$. Then $((1,0),n)\in\pi_0\circ\Psi$, hence $((1,0),n)\in F_0\circ\pi_0$. Then $(1,n)\in F_0$ a contradiction.
A sufficient condition so that this set exists is that for every $a\in A$ there are $b,c\in A$ such that $(a,b)\in F_0$ and $(a,c)\in F_1$. Indeed if that is the case let $$\Psi=\{(a,b,c,d) : (a,c)\in F_0\land (b,d)\in F_1\}.$$ If $(a,b,c)\in\pi_0\circ\Psi$ then there is some $d$ such that $(a,b,c,d)\in\Psi$, hence $(a,c)\in F_0$ and thus $(a,b,c)\in F_0\circ\pi_0$. On the other hand if $(a,b,c)\in F_0\circ\pi_0$ then $(a,c)\in F_0$. Since by the assumption there is some $(b,d)\in F_1$, $(a,b,c,d)\in\Psi$ and hence $(a,b,c)\in\pi_0\circ\Psi$. In fact this is also a necessary condition - to see this notice how the above counterexample worked.
I may later add some stuff regarding the uniqueness.
Edit: Even if this set exists it needs not be unique. The counterexample is the following: Let $A=2$, $F_0=F_1=\{(1,0),(1,1),(0,0)\}$ and $\Phi=\varnothing$. Then one can notice that if $\Psi$ is the set I give above then $\Psi'=\Psi\setminus\{(1,1,1,1)\}$ also works.