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Let there be three relations:

$g\subseteq D\times A\\ h\subseteq D\times B\\ R\subseteq(A\times B)\times C$

For the purposes of this post, the "composite of $R$ with $g$ and $h$," a subset of $D\times C$, is defined as

$R_{g,h}=\{(d,c)\in D\times C\mid \exists (d,x)\in g, (d,y)\in h \text{ such that } ((x,y),c)\in R\}$

I know the category Rel of binary relations uses sets as objects and relation as arrows, and relation composition (in the sense described here) as composition. Otherwise I have not worked in it, and to be honest it's a little bit weird.

I keep feeling like what's being described above is some simple categorical construct interpreted in Rel (like a pushout or some product), but it's hovering just outside of my grasp.

I'm faced with the given three arrows above, and the final arrow constructed with them, and I try to organize them in a diagram.

There should be an arrow from $D\to A\times B$ making a commutative diagram connecting $D,A,B$ and $A\times B$. Then $R$ gives this arrow to $C$, then naively I would look at the composition of arrows from $D$ to $A\times B$ to $C$, but I'm not sure this diagram is useful at all.

My two specific questions are:

  1. What's the interpretation of this composition categorically (if there is one)?

  2. Does it have a less bland name in the literature somewhere? (I would like to search for it.)

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If I'm not mistaken, the relation $R_{g,h}$ is just $R\circ (g,h)$, where $(g,h) = g\times h \circ (\operatorname{id}_D, \operatorname{id}_D)$ is a relation from $D$ to $A\times B$.

The function / morphism $(\operatorname{id}_D, \operatorname{id}_D)$ is identified with its graph for the purpose of composing relations and $\times$ is just the Cartesian product applied to relations. So this makes sense categorically as well.

I don't know whether this has any special name.

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  • $\begingroup$ Tell me if you need more details about the categorical constructions involved. $\endgroup$
    – Stefan Perko
    Oct 10, 2017 at 15:34
  • $\begingroup$ OK, then I wasn't totally off when I suggested this composition. Thank you for help putting things in perspective. You might be interested in the one example that was given: $f_{I, w}$ where $I$ is the identity function on $\mathbb R$, $w$ is the function which is constantly $w$ on $\mathbb R$, and $f$ is a differentiable function $\mathbb R\to \mathbb R$. Look familiar? It appears in every calculus book, but definitely not in this guise... $\endgroup$
    – rschwieb
    Oct 10, 2017 at 16:29
  • $\begingroup$ @rschwieb Sorry, I don't get your example. With what kind of relations do you identify $f$ and $w$? If $w : \mathbb R \to \mathbb R$, then $f$ should be $\mathbb R^2 \to \mathbb R$. - In any case it is probably a good idea to put the example into the question to provide more context. $\endgroup$
    – Stefan Perko
    Oct 10, 2017 at 16:41
  • $\begingroup$ Doh, you're right, I see what you mean. I haven't given $R$ correctly. Hang on... $\endgroup$
    – rschwieb
    Oct 10, 2017 at 17:17
  • $\begingroup$ Given a differentiable function $f$, use the relation $R^f(x,y)=\frac{f(x)-f(y)}{x-y}-f'(y)$ when $x\neq y$ and $R^f(x,x)=0$. Then compose $R^f_{I, w}$. Maybe not as amusing as I first found it. It's certainly a weird way to look at the difference quotient. $\endgroup$
    – rschwieb
    Oct 10, 2017 at 17:22

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