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?

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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