Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

In section III.1 of P.M. Cohn's Universal Algebra a notion of universal functor ${\cal L} \rightarrow {\cal K}$ is defined for a representation of one category in another given by a (covariant) functor $F: {\cal L}^{\mbox{opp}} \times {\cal K} \rightarrow \mbox{Set}$. The objects part of the functor is specified by a universal property and Cohn states and proves a proposition (1.1) stating that if this exists, then it is indeed the objects part of a functor $U$ and that a certain mapping $\rho$ is a natural transformation from $I$ to $U$.

Cohn writes $\rho$ as $\rho(A) : A \rightarrow U(A)$ where $A \in \mbox{Ob} {\cal L}$. But as $A$ and $U(A)$ are objects in different categories, even given his convention of omitting certain "obvious" functors, I can make no sense of this in the general case, where the representation is not given by a forgetful functor. (To make things "type correct", $\rho(A)$ has to be an element of the set $F(A, U(A))$).

I can see how to define the morphisms part of the functor $U$ using the universal property and how to prove that $U$ is then a functor by a diagram chase in $\mbox{Set}$, but this is nothing like Cohn's proof which appears to compose morphisms from two different categories. I can also see how Cohn's proof would say something in the case where the representation is derived from a forgetful functor, but I can't even see how to formulate the claim that $\rho$ is a natural transformation from $I$ to $U$ in the general case. I doubt very much that Cohn is wrong, so what am I missing?

share|improve this question

2 Answers 2

What Cohn calls a representation is what Jacob Lurie calls a correspondence in Higher Topos Theory (which you can download from his website), in a subsection called "correspondences", which is 2.3.1 in my copy. Given a correspondence $C'^{op}\times C\rightarrow Set$ we can form the join along this correspondence, as you may read in Lurie's book. Then take "I" to be the inclusion into this join, and "U" to be as described in Cohn.

share|improve this answer

I have the 1965 edition of Cohn's book, which apparently differs from the one you cite, since what you refer to as Proposition 1.1 is Theorem 1.2 in my copy. Nevertheless, I hope the following comments may be useful.

First, about the mysterious "morphisms" between objects that live in different categories, Cohn writes (near the top of page 110 in my copy) "we generally refer to the elements of $F(A,a)$ as the admissible morphisms of the representation. I believe this explains the anomaly that bothered you.

Let me also mention that this overloading of the word "morphism" is not as silly as it might at first appear. There is a natural way to "glue together" two categories $\mathcal L$ and $\mathcal K$ using a functor $F$ as in your question. Start with the disjoint union of $\mathcal L$ and $\mathcal K$, and then add Cohn's admissible morphisms as actual morphisms from an object of $\mathcal L$ to one of $\mathcal K$. Of course, having added these morphisms, you need to define compositions involving them. But $F$ tells you how to do that: If $\rho:A\to a$ is one of the new morphisms, you want to pre-compose it with morphisms $\lambda:B\to A$ of $\mathcal L$ and to post-compose it with morphisms $\kappa:a\to b$ of $\mathcal K$. For the former, act on $\rho$ by $F(\lambda,1_a)$; for the latter, use $F(1_A,\kappa)$. The functoriality of $F$ ensures that this definition of composition for the new morphisms plus the pre-existing composition laws of $\mathcal L$ and $\mathcal K$ satisfy the associative law.

I conjecture that this construction is the "join" mentioned in Herman Stel's answer, but I haven't checked "Higher Topos Theory" to make sure. I do, however, assure you that this construction was the mental picture of $F$ that came naturally to my mind when I started reading your question.

share|improve this answer
    
Yes, this is the join in Lurie's book. –  Herman Stel Aug 16 '10 at 7:28
    
The Google Books link I gave is for the 2nd edition of Cohn. Thanks to you both for the very helpful answers. This seems like quite a wide lacuna for the unsuspecting reader to me. –  Rob Arthan Aug 25 '10 at 21:56

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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