products in a category without reference to objects or sources and targets

Hi,

I was thinking about presenting categories with nothing but equations over morphisms. I wondered about products. The definition of a product has its genesis in the following diagram shape

A->B

A->C

You would say that whenever you have this shape, then for any D such that

D->A->B

and

D->A->C

blah blah...the axioms for the product. I thought about how to do this with just words in the arrows without reference to either source and target or objects. You can start with words like

$dab$ and $dac$

where a:D->A and and b: A->B and c:A->C and d:E->D.

You might say: if, the existence of the equations

$dab=e$ and

$dac=f$

implies that $a$ is unique in that if there exists words

$dxb$ $dyc$

then $x=y=a$

then this constitutes a product.

Could something like this work? I mean, can you present a category as just words over morphisms along with equations and a bit of language with quantifiers? Second, could this kind of definition work for products?

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I don't understand what you mean by "without reference to objects or sources and targets." If you understand how to compose morphisms, that's the same thing as understanding objects, sources, and targets. –  Qiaochu Yuan Jun 25 '11 at 4:30
Qiaochu: You mean "understand which pairs of morphisms can be composed", right? –  Tom Goodwillie Jun 25 '11 at 10:22
Qiaochu, By "the same thing" I think we would both mean that two different presentations are actually presenting theories of the same "thing" ie the theory of categories. Recall, I am presenting a category as a long list of words and equations. In this kind of presentation, you "know" that two morphisms, $a,b$ can be composed if anywhere in the list of you see a word with a subword $ab$. You know two words $ab$, $cd$ can be composed if anywhere in the list of equations there is a word with subword $bc$. –  Ben Sprott Jun 25 '11 at 11:43
@Ben: you can't present a category as just words and equations. For example, here are two categories presented normally: C has two objects x and y, and Hom(x,y) has two arrows f and g; D has four objects a, b, x, y and Hom(a,b) and Hom(x,y) each have a single arrow f, g respectively. In your language, there are only two letters, namely f and g, no other words, and no equations in either category. So how do you tell C from D? –  Ryan Reich Jun 25 '11 at 13:32
@Ben: (continued) You might argue that I forgot to mention the identity arrows. That's true; if I mentioned them, it would be exactly the same as just telling you what all the objects are. Then the "words" that you want, which express composability, would tell me what the source and targets are for each arrow (just check which identity arrows something is composable with). There is no escaping it. –  Ryan Reich Jun 25 '11 at 13:34

As Freyd and Scedrov show in "Categories, Allegories", you can think of a category as some kind of partial monoid. Such a monoid has a set of partial units that take over the role of objects in standard presentations of categories.

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I see all this as an exercise in foundations. I guess I am looking for a very lean presentation of the theory of categories. I think this partial monoid thing looks pretty lean. –  Ben Sprott Jun 26 '11 at 14:13

Yes this is possible. But you have to restrict to "composable" words.

A category may be viewn as a monoid whose composition law is only defined partially: We have a set of morphisms $M$ and maps $s,t : M \to O$ (source/target), then $mn$ is defined as soon as $t(n)=s(m)$ and satisfies $t(mn)=t(m), s(mn)=s(n)$. But this "algebraic point of view" is often not so enlightening. It is better to picture your category as sort of a graph with a notion of commutative diagram. The product of two objects $p,q$ is the "smallest" object $p \times q$ which lies "above" $p,q$ in the sense the there are two arrows $p \times q \to p$ and $p \times q \to q$. You should really draw this.

Just to compare, here is the algebraic definition: A product $p \times q$ is a pair of $m,n \in M$ with $s(m)=s(n), t(m)=p, t(n)=q$, such that for each other such pair $(m',n')$, there is a unique $u \in M$ such that $s(u)=s(m'), t(u)=s(m)$ and $m'=mu, n'=nu$.

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I think I see lots of references to objects, source and targets here :) –  Theo Buehler Jun 25 '11 at 5:26
Well we all know (I hope, Ben too) that we cannot just throw away the objects. I just wanted to indicate that it is "syntactical" possible what Ben was asking for. –  Martin Brandenburg Jun 25 '11 at 7:33
Seconding Martin B's answer/comment: yes, it is possible to formulate "category theory" linguistically so as to not literally refer to objects. Unsurprisingly, (as in Wouter S.'s answer, and in comments above) the "partial units" recover the objects. It is a bit of an exercise in circumlocution, and, seriously, I think it proves, as much as anything, that it is possible to obscure operational intentions while nevertheless being logically correct, even "logically economical". –  paul garrett Jul 24 '11 at 13:06