Double embedding or double restriction

Question

The following generalizes both embedding and restriction for sets $A$ and $B$:

$A \rightleftarrows B = ( A ; B ; \operatorname{id}_{A \cap B})$.

$A \rightleftarrows B$ is considered as a morphism of the category $\mathbf{Rel}$.

It is an embedding $A \hookrightarrow B = ( A ; B ; \operatorname{id}_A)$ if $A\subseteq B$.

It can be called a restrition if $B\subseteq A$ (in this case $A \rightleftarrows B = ( A ; B ; \operatorname{id}_B)$.).

It is well known that embedding into a set $A$ and then embedding into a set $A'$ is the same as embedding into $A'$ (if $A\subseteq A'$). In simple words: Double embedding is embedding.

It is well known that restricting to a set $A$ and then restricting to a set $A'$ is the same as restricting to $A'$ (if $A'\subseteq A$). In simple words: Double restriction is restriction.

The following obvious theorem (enough conditions for $( B \rightleftarrows C) \circ ( A \rightleftarrows B) = ( A \rightleftarrows C)$) generalizes it:

1. $( B \rightleftarrows C) \circ ( A \rightleftarrows B) = A \rightleftarrows C$ if $C \subseteq B$.
2. $( B \rightleftarrows C) \circ ( A \rightleftarrows B) = A \rightleftarrows C$ if $A \subseteq B$.

(Here composition of $\mathbf{Rel}$-morphisms is induced by the composition of binary relations: $g\circ f=\{(x;z) \,|\, \exists y:(xfy\wedge ygz) \}$.)

My question: What are necessary conditions for $( B \rightleftarrows C) \circ ( A \rightleftarrows B) = ( A \rightleftarrows C)$?

My guess the condition is $A\subseteq B\vee C\subseteq B$.

Note: The above is for the category $\mathbf{Rel}$. I use it in my research of categories $C$ equipped with a functor $\mathbf{Rel}\rightarrow C$. Thus it is important.

Answer 1

$( B \rightleftarrows C) \circ ( A \rightleftarrows B) = A \rightleftarrows C$ is equivalent to:

$(B;C;\operatorname{id}_{B\cap C})\circ(A;B;\operatorname{id}_{A\cap B}) = (A;C;\operatorname{id}_{A\cap C})$;

$(A;C;\operatorname{id}_{A\cap B\cap C}) = (A;C;\operatorname{id}_{A\cap C})$;

$A\cap B\cap C = A\cap C$;

$B\supseteq A\cap C$.

So the answer is $B\supseteq A\cap C$.