# Adjunction Signature: abuse of notation or actual functor?

In "Relational Algebra by Way of Adjunctions," found at author's page (doi), section 2.4, an adjunction is described using the signature:

$$L \dashv R:\mathscr{D}\to \mathscr{C}.$$

Based purely on how my understanding of type signatures work, the above states that the concept, $$L \dashv R$$, is an arrow from $$\mathscr{D}$$ to $$\mathscr{C}$$--a Functor in this case. However, it seems to me that this may be a convenient (and seemingly standard) way to call out the relevant categories involved. From the adjunctions in the paper, $$\eta$$ and the component functors are utilized, but I don't see that $$L \dashv R$$ is ever actually used as a $$\mathscr{D} \to \mathscr{C}$$ functor in its own right.

On the other hand, I'm in no position to just assume that the authors didn't really mean what they wrote--these guys are good. So my question, is $$L\dashv R:\mathscr{D}\to \mathscr{C}$$ a functor? What is the definition of that functor for $$\Delta\dashv\times$$, or any of the adjunctions in figure 3?

• Section 2.4 of that paper clearly states that $L$ and $R$ are a pair of functors. – John Douma Jan 25 '19 at 15:54
• @JohnDouma, yes, that much is clear. $R \circ L : C \to C$ is also a functor. The $\circ$ makes a new functor from old. The question essentially asks does $\dashv$ make a new functor from old in a similar way. – trevor cook Jan 25 '19 at 16:13
• What do you mean by a similar way? – John Douma Jan 25 '19 at 16:34
• @JohnDouma, just that they both look like binary operators. – trevor cook Jan 25 '19 at 18:07

## 2 Answers

No, $$L\dashv R$$ is a potentially confusing shorthand for "$$L: C\to D,R:D\to C$$, and $$L$$ is left adjoint to $$R$$."

You could rationalize the notation as $$L\dashv (R : \mathcal D \to \mathcal C)$$ where we're ascribing a "type" to $$R$$ for clarity, but really $$L\dashv R : \mathcal D \to \mathcal C$$ is just treated as a (common) short hand for $$L\dashv R$$ and $$R:\mathcal D \to \mathcal C$$. There's nothing deeper going on here other than compactly indicating the relevant categories. $$L\dashv R$$ is typically viewed as a proposition asserting that $$L$$ is left adjoint to $$R$$ and so is a formula not a term.

That said, there is a (2-)category of adjunctions and (over $$\mathbf{Cat}$$) its objects are categories. You could also rationalize the above as stating the adjunction $$L\dashv R$$ is an arrow in that 2-category. You seem to have the misapprehension that the objects of a category determine the category. It is quite possible to have multiple categories with the same collection of objects. If $$X$$ and $$Y$$ are objects, you don't automatically know what an arrow $$X\to Y$$ is. You need to know in which category you're working. The fact that there's some category with $$X$$ and $$Y$$ as objects that you're familiar with doesn't mean there isn't some other category which also has $$X$$ and $$Y$$ as objects but has a different notion of arrows. For example, $$\mathbf{Set}$$ is the category of sets and functions, but $$\mathbf{Rel}$$ also has sets as objects but relations as arrows.

• I get your point, but you have the direction of $R$ reverse of the paper's. Does that change the rationalized notation? Also,the 2-cat of adjunctions is interesting, is that where the notation came from? – trevor cook Jan 25 '19 at 19:54
• Ugh. Well, that convention would be consistent with the arbitrarily suggested convention on the nLab page for the 2-category of adjunctions. I'm pretty sure I usually see it the other way, e.g. as indicated by Kevin Carlson's answer. I don't know if $\mathsf{Adj}(\mathbf{Cat})$ is the origin of the notation. I suspect it is not. It certainly isn't the explicit justification for the notation most of the time it is used since the 2-category of adjunctions isn't talked about that often. – Derek Elkins left SE Jan 25 '19 at 20:57
• Thanks. Wish i could accept this too. – trevor cook Jan 26 '19 at 13:00