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Given two relations $R\subseteq A\times B$ and $R'\subseteq A'\times B'$. Is it known/used that every relation $r\subseteq R\times R'$ can be characterized by two relations $\alpha\subseteq A\times A'$ and $\beta\subseteq B\times B'$ so that

$((a,b),(a',b'))\in r \iff \Big((a,a')\in\alpha\wedge (b,b')\in\beta\wedge (a,b)\in R\implies (a',b')\in R'\Big)$

and if $R''\subseteq A''\times B''$, $\,r'\subseteq R'\times R''$, where $r'$ is characterized by $\alpha'\subseteq A'\times A''$ and $\beta'\subseteq B'\times B''$, then the composition $r'\circ r$ is characterized by the relations $\alpha'\circ\alpha\subseteq A\times A''$ and $\beta'\circ\beta\subseteq B\times B''$? (Where $\circ$ denote the composition of relations). $\;$In diagram form: $\require{AMScd}$ \begin{CD} A@>\alpha>>A'@>\alpha'>>A''@. A@>\alpha'\circ\alpha>>A''\\ @VRVV r @VVR' V r'@VVR''V \quad\implies\quad @VRVV r'\circ r @VVR''V\\ B@>>\beta>B'@>>\beta'>B'' @. B@>>\beta'\circ\beta>B'' \end{CD} Does this category of relations as objects and relation between relations as morphisms have a name?


The reason for my interest in this category:

Suppose $A=B\times B$ and that $R\subseteq A\times B$ is the composition in a magma. Then the functions among the morphisms between two such objects defines magma morphisms $B\to B'$.

Suppose $B=\mathcal P(A)$ and that $R\subseteq A\times B$ is the relation $(a,S)\in R\iff a\in\overline{S}$ for some topology on $A$. Then the functions among the morphisms between two such objects define continuous functions $A\to A'$.


Edit:
The category of relations as objects and pairs of relations defined by some $\alpha,\beta$ (as above) as morphisms, is a subcategory of the category of relations as objects and relations of relations as morphisms. Very few relations between relations can be defined with two relations $\alpha,\beta$, but those relations which can seems to be an interesting category.

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  • $\begingroup$ @Chilango, see math.stackexchange.com/questions/1857985/… $\endgroup$
    – Lehs
    Commented Jul 17, 2016 at 16:34
  • $\begingroup$ I know what $Rel2$ is I'm just wondering what you want to do with the above category, or you just want a name for it? Cheers. $\endgroup$ Commented Jul 17, 2016 at 16:43

1 Answer 1

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Such a characterisation of 'relations between relations' is not possible.

As a counterexample, let $A,B,A',B',\alpha,\beta$ be whatever you want them to be, let $R = \varnothing$, let $R' = A' \times B'$. Then:

  • $(a',b') \in R'$ for all $a' \in A'$ and $b' \in B'$, so the statement on the right-hand side of your $\Leftrightarrow$ symbol is true for all $a,b,a',b'$.
  • $R \times R' = \varnothing$, so the only relation $r \subseteq R \times R'$ is the empty relation. Hence the left-hand side of your $\Leftrightarrow$ symbol is false $a,b,a',b'$.

I can't answer your other questions, since they are contingent on such a characterisation of 'relations between relations' being possible.

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  • $\begingroup$ Didn't you miss the $(a,b)\in R$ part on the right side of the $\iff$? $\endgroup$
    – Lehs
    Commented Jul 17, 2016 at 17:26
  • $\begingroup$ @Lehs: For an implication $p \Rightarrow q$ to be true, it suffices for $q$ to be true. In this case, $(a',b') \in R'$ is always true, so the stuff to the left of the $\Rightarrow$ symbol doesn't affect the truth of the whole implication. So even though $(a,b) \in R$ is always false, the implication on the right-hand side of the $\Leftrightarrow$ is always true. $\endgroup$ Commented Jul 17, 2016 at 17:29
  • $\begingroup$ Wording this in a slightly clearer way: the statement $$(a,a') \in \alpha \wedge (b,b') \in \beta \wedge (a,b) \in R \Rightarrow (a',b') \in R'$$ is true provided that $(a',b') \in R'$ is true, which is the case when $R' = A' \times B'$, no matter what values $a,b,a',b'$ take. $\endgroup$ Commented Jul 17, 2016 at 17:30
  • $\begingroup$ Yes, putting it that way I can see that the conjecture is false and that the relations characterized by some $\alpha,\beta$ is a proper subset of all relations (between relations). Thanks! This has confused me for years and it seems that some work is still to be done. $\endgroup$
    – Lehs
    Commented Jul 17, 2016 at 17:43

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