1
$\begingroup$

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?

$\endgroup$
4
  • $\begingroup$ Section 2.4 of that paper clearly states that $L$ and $R$ are a pair of functors. $\endgroup$ – John Douma Jan 25 '19 at 15:54
  • $\begingroup$ @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. $\endgroup$ – trevor cook Jan 25 '19 at 16:13
  • $\begingroup$ What do you mean by a similar way? $\endgroup$ – John Douma Jan 25 '19 at 16:34
  • $\begingroup$ @JohnDouma, just that they both look like binary operators. $\endgroup$ – trevor cook Jan 25 '19 at 18:07
5
$\begingroup$

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

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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? $\endgroup$ – trevor cook Jan 25 '19 at 19:54
  • 1
    $\begingroup$ 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. $\endgroup$ – Derek Elkins left SE Jan 25 '19 at 20:57
  • $\begingroup$ Thanks. Wish i could accept this too. $\endgroup$ – trevor cook Jan 26 '19 at 13:00

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.