3
$\begingroup$

Probably a silly question. Suppose that $C$ is a category that does not have finite Cartesian products. So we cannot define a relation on some objects to be a sub object of their Cartesian product (a monic arrow into their Cartesian product). Is there some other natural notion that we can use $inside$ the category to generalise the notion of `relation'? I'm not interested in using the concretisation, so let's suppose $C$ is not concrete.

$\endgroup$
9
$\begingroup$

You could describe a relation between $X$ and $Y$ to be a pair of maps $f\colon R\to X$, $g\colon R\to Y$, so that the family of maps $\{f,g\}$ is monic (meaning, if $fh=fh'$ and $gh=gh'$, then $h=h'$.)

$\endgroup$
  • 4
    $\begingroup$ One also says "$f$ and $g$ are jointly monic"... $\endgroup$ – მამუკა ჯიბლაძე Oct 14 '14 at 13:38
  • $\begingroup$ In fact, the above is the standard definition of an internal relation: an internal relation is a span of morphisms that are jointly mono. In case the category has binary products such relations can be represented as single monomorphisms into cartesian products. The latter definition is a bit less convenient to work with, but it is much easier to generalize it to non-canonical notions of subobjects, therefore some authors use it as their primary definition. $\endgroup$ – Michal R. Przybylek Oct 14 '14 at 14:24
  • 2
    $\begingroup$ All this being said, it's hard to do much with relations in a category $C$ to simulate the usual sort of calculus of relations, unless one assumes more of $C$. For example, to get a half-decent notion of composition of relations, one typically assumes that $C$ is a regular category. $\endgroup$ – Todd Trimble Oct 14 '14 at 14:28
  • $\begingroup$ @ToddTrimble, but, actually, you do not have to assume that $C$ has products to define associative compositions (it suffices to assume that $C$ has pullbacks and stable images). $\endgroup$ – Michal R. Przybylek Oct 14 '14 at 16:15
  • $\begingroup$ Of course not, @MichalR.Przybylek $\endgroup$ – Todd Trimble Oct 14 '14 at 17:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.