1

Objects in the category Rel2 (my notation) are the relations $r\subseteq A\times B$, $r'\subseteq A'\times B'$ (the morphisms in Rel) and morphisms are pair of relations $\alpha\subseteq A\times A'$ and $\beta\subseteq B\times B'$ such that

$(1)\quad(a,a')\in\alpha\wedge(b,b')\in\beta\implies\big((a,b)\in r\implies(a',b')\in r'\big)$

or equivalently

$(1')\!\!\quad(a,a')\in\alpha\wedge(b,b')\in\beta\wedge(a,b)\in r\implies(a',b')\in r'$.

My question is if all relations $R\subseteq r\times r'$ define a morphism in Rel2, that is, if given $R$ there exist relations $\alpha\subseteq A\times A'$ and $\beta\subseteq B\times B'$ such that:

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

Some context:

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:
It is certainly two different categories. Suppose all four sets $A,B,A',B'$ have two elements. Then the number of relations between two such sets is $2^{2\times 2}=16$ and therefore the number of different pairs $\alpha,\beta$ is $16\times 16=256$.

While the number of relations between two relations is $2^{16\times 16}=$ $115792089237316195423570985008687907853269984665640564039457584007913129639936$

Community
  • 1
Lehs
  • 13,791
  • 4
  • 25
  • 77
  • The condition you've written implies that $R$ is all of $r\times r'$, since if $(a',b')\in r'$ then the implication is automatically true. – Eric Wofsey Jul 14 '16 at 18:41
  • Can you write the first section a bit more clearly? It looks interesting, but I can barely understand what's going on... – goblin GONE Jul 18 '16 at 04:16
  • @goblin, it's also formulated in http://math.stackexchange.com/questions/1862245/does-this-category-have-a-name – Lehs Jul 18 '16 at 04:22
  • Its still not clear to me, but let me emphasize that this looks super cool and that I'd encourage a careful rewrite. – goblin GONE Jul 18 '16 at 04:42
  • @goblin, I can't work it out in this question, but I will try to explain this category in my profile, as soon as I can. – Lehs Jul 18 '16 at 04:50

1 Answers1

1

$\require{AMScd}$I don't quite understand the question, but basically, I think you're trying to define a particular double category.

Objects. Sets

Arrows. Relations

Proarrows. Relations

Squares. We assume that each square has at most one filler, and that it has a filler iff the condition $$\quad(a,a')\in\alpha\wedge(b,b')\in\beta\implies\big((a,b)\in r\implies(a',b')\in r'\big)$$ is satisfied.

(Think of $\alpha$ and $\beta$ as arrows, and $r$ and $r'$ as pro-arrows.)

This looks like a potentially important idea, and I'd encourage you to keep thinking more about it. You may or may not find this post of mine relevant, or at least interesting.

Community
  • 1
goblin GONE
  • 67,744
  • Yes, something is interesting in this, but i'll never studied 2-categories. The construction reminds about morphisms between directed graphs. – Lehs Jul 18 '16 at 04:54
  • @Lehs, you should have a look at double categories, since they're very fundamental. – goblin GONE Jul 18 '16 at 04:55
  • I know the concept of squares as morphisms from homological algebra, but I studied category theory for the purpose of traditional mathematics and I have no clue what multi categories are for. – Lehs Jul 18 '16 at 05:10
  • @Lehs, the purpose of higher-category theory is traditional mathematics! The same can be said of multicategories, by the way. You do realize that $\mathbf{Ab}$ is a multicategory, right? So is the (multi)category of medial magmas – goblin GONE Jul 18 '16 at 07:16
  • How is Ab a multicategory? I'll ask a question about multicategories. – Lehs Jul 18 '16 at 07:26
  • @Lehs, abelian groups and multilinear maps form a multicategory. – goblin GONE Jul 18 '16 at 07:39
  • I would appreciate if you elaborate on this theme here http://math.stackexchange.com/questions/1862995/multicategories-for-their-own-sake-or – Lehs Jul 18 '16 at 14:36
  • @Lehs, okay. But I have an exam tomorrow that I have to study for. Maybe tomorrow or the day after I'll write an answer. – goblin GONE Jul 18 '16 at 15:51
  • I look forward to that. – Lehs Jul 18 '16 at 16:18
  • @Lehs Any monoidal category $\mathcal{C}$ yields a multicategory by letting the multimorphisms $(c_1, \dots, c_n) \to d$ be the morphisms $c_1 \otimes \dots \otimes c_n \to d$. See this explanation. – Najib Idrissi Jul 19 '16 at 11:52
  • @NajibIdrissi, in principle that's correct, but pedagogically, I think its backwards. Better to first introduce multilinear maps first (e.g. to the 1st year undergraduate linear algebra class), and then introduce the tensor product second, explaining that it represents multilinear maps (e.g. to the 1st year graduate algebra class). – goblin GONE Aug 05 '16 at 06:38