MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm looking for a reference for a certain pair of monads on $Cat$. One problem is that I don't know the modern way of thinking about some basic things, so excuse me if my presentation is naive.

First some notation. Let $C$ and $D$ be small categories and let $F\colon C\to Cat\;$ and $G\colon D\to Cat\;$ be functors. Let $(F\Uparrow G)$ denote the following category. $$Ob(F\Uparrow G)=\{(c,d,\ell)\;|\;c\in Ob(C), d\in Ob(D), \ell\colon F(c)\to G(d)\}$$ and whose morphisms are ``natural transformation squares", i.e. $$Hom_{(F\Uparrow G)}((c,d,\ell),(c',d',\ell')=\{(f,g,\alpha)\;|\;f\colon c\to c', g\colon d\to d', \alpha\colon G(g)\circ\ell\to\ell'\circ F(f)\}.$$ With the same setup, let $(F\Downarrow G)\;$ denote the category with the same objects but slightly different morphisms, the only difference being that the natural transformation point the other way: $$Ob(F\Downarrow G)=\{(c,d,\ell)\;|\;c\in Ob(C), d\in Ob(D), \ell\colon F(c)\to G(d)\}$$ $$Hom_{(F\Downarrow G)}((c,d,\ell),(c',d',\ell')=\{(f,g,\beta)\;|\;f\colon c\to c', g\colon d\to d', \beta\colon\ell'\circ F(f)\to G(g)\circ\ell\}.$$

Let $D=\{\star\}$, and denote a functor $G\colon D\to Cat\;$ by $\{G\}$. Now let $C=FCat$, some skeleton of the category of finite categories. Then we have a functors $$(FCat\Uparrow\{-\})\colon Cat\to Cat \;\;\;\;\text{ and }\;\;\;\;(FCat\Downarrow\{-\})\colon Cat\to Cat.$$ I think that each is the functor part of some kind of 2-monad on $Cat$. The unit is "constant" and the multiplication is "Grothendieck construction".

Proving that this is associative, etc, looks laborious, and I don't want to reinvent notation, etc. Is there a good reference for these monads, if they really are monads?


share|cite|improve this question
up vote 8 down vote accepted

What you are describing is an example of Max Kelly's notion of club, closely connected with the concept of operad. The original references date back to the 70's; one reference is

  • G.M.Kelly. On clubs and doctrines. In Category Seminar, Sydney 1972/1973. Springer LNM 420, pp. 181-257 (1974).

Actually, a club is defined to be a monoid in a monoidal category $M$ whose objects are pairs $(C, F: C \to Cat)$ where $C$ is a small category, and where a morphism $(C, F) \to (D, G)$ consists of a functor $H: C \to D$ and a natural transformation $G H \to F$; the monoidal product is a kind of wreath product. There is an "actegory" structure (an action of the monoidal category $M$) on $Cat$,

$$\wr: M \times Cat \to Cat,$$

so that each club = monoid in $M$ induces a monad on $Cat$. You can also find a succinct description of this material here.

As a reality check, here is a more direct description of the (underlying functor of the) monad on $Cat$, associated with the club structure on the inclusion $i: \mathrm{FCat} \hookrightarrow Cat$, which can be extracted by applying the construction given in Borisov's paper, pp. 3-4. The monad takes a category $D$ to a category which I will denote $\mathrm{FCat} \wr D$ (Borisov uses a semi-direct product symbol). The objects of $\mathrm{FCat} \wr C$ are pairs $(C, l: i(C) \to D)$ where $C \in \mathrm{FCat}$ is a finite category and $l: i(C) \to D$ is a functor. Morphisms $(C, l) \to (C', l')$ of $\mathrm{FCat} \wr D$ are pairs $(F: C \to C', \phi: l \to l' \circ i(F))$ where $\phi$ is a natural transformation.

Unless I have misunderstood something, this gives one of the monads described in the OP. I think the other monad is obtained by a process of dualization (apply $(-)^{op}$, then apply the first monad, then apply $(-)^{op}$ again), so we really only need to worry about the first. As I said, all the laborious technical details were covered long ago in the "Australian school".

Other familiar examples of clubs are where we take the inclusion $i: Set \to Cat$, mapping each set $S$ to the discrete category on $S$; the corresponding monad is the free coproduct completion monad. Another is the composite inclusion $\mathbb{P} \hookrightarrow Set \hookrightarrow Cat$ of the groupoid of finite permutations; here the corresponding monad is the free symmetric (strict) monoidal category construction.

share|cite|improve this answer
I don't see the connection exactly. In my notation, you seem to be talking about $(id_{Cat}\Downarrow \{Cat\})$. Am I missing something? .........<newParagraph>.......... Suppose I have categories $C,D$ and a functor $F\colon C\to (FCat\Downarrow \{D\})$, i.e. for every object in $C$ a finite category over $D$ and for every morphism in $C$ a natural transformation triangle. I need to receive a finite category over $D$, i.e. an object in $(FCat\Downarrow \{D\})$. Can you relate you're talking about to this situation? – David Spivak Feb 20 '13 at 22:34
@David: I have quickly browsed through the article by Dennis Borisov linked in Todd Trimble's answer. I think the club that Todd is implying gives your example is a monoid in $M$ (the category in the above answer) whose underlying object is the category $\mathrm{FCat}$ with its inclusion into $\mathrm{Cat}$. There definitely seems to be a very close relation between that club and your question. Unfortunately, there are many details in the description of clubs (as described by Dennis Borisov's article), and it is not yet clear to me that your case is exactly an instance of that framework. – Ricardo Andrade Feb 21 '13 at 0:07
@David: Sorry for the delay in responding; I am on vacation, and needed a spare moment to write out something further, which I have done in an edit. I hope I understood you correctly. – Todd Trimble Feb 21 '13 at 22:17
Thanks for clarifying. Now I understand what Borisov is doing there! This was very helpful and actually answers an old mathoverflow question of mine, "the urge to combine 1- and 2-morphisms in slicing a 2-category", and I realize that Finn Lawler understood that. – David Spivak Feb 25 '13 at 0:18
Okay, great! Glad it helped. – Todd Trimble Feb 25 '13 at 1:48

Your Answer


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.