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The limit of a functor $F:D^{op}\to C$ is an object $\lim F$ representing the functor $$\ell F(x):=\operatorname{Psh}_D(\ast,C(x, F(\cdot))),$$ where $*$ denotes the terminal presheaf on $D$. (Notice that $C(x, F(\cdot)))$ is a presheaf on $D$).

We can define the limit of a functor weighted by a presheaf in much the same way (by replacing $\ast$ with a fixed presheaf on $D$ called the weight).

Why am I bringing this up? It is a very slick definition. Nowhere do we have anything like universal arrows popping up. Adjunctions are out of sight and out of mind. Indeed, this definition generalizes straightforwardly to S-enriched categories for S symmetric monoidal closed (and all of the other requirements you need for the S-enriched Yoneda lemma to work).

The usual definition of the Kan extension is as a functor completing a certain commutative triangle such that it is universal in a specific sense in a certain functor category (intentionally vague...). This definition is pretty annoying to work with and is avoided whenever possible by instead insisting that all Kan extensions be pointwise (for instance, Kelly does this his book on enriched categories).


Does there exist a similar slick definition of the Kan extension (not necessarily pointwise)? By slick here, we mean free of adjoint functors (and their less conspicuous cousins, universal arrows) and free of commutative diagrams (translating the content of a commutative diagram into prose does not count).

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I'd love to see your displayed formula! –  Jeff Strom Jul 24 '10 at 12:16
@Jeff: I don't follow. What do you mean? –  Harry Gindi Jul 24 '10 at 12:35
I'm guessing that something (maybe the asterisk?) in your formula for the functor represented by the limit of $F$ prevented it from appearing as intended when Jeff viewed it. Sometimes a question or answer here fails to "TeX" properly for me (but not this time). Oddly, when this happens to me I can usually view the thing better by making as if I'm going to edit it. –  Tom Goodwillie Jul 24 '10 at 14:33
@Harry I can't view it properly either. I just see some raw looking TeX. Maybe replace * with \ast ? –  Dan Piponi Jul 24 '10 at 16:59

1 Answer 1

up vote 7 down vote accepted

Firstly, the $W$-weighted limit $\lim^W F$ is defined to be a representation of $\operatorname{Psh}_D(W,C(-,F-))$; the definition you've given isn't even well-typed.

There is no difference at all in $\mathrm{Set}$-enriched category theory between a representation and a universal arrow -- each determines the other, and when these exist for all suitable objects then adjoints are there whether you like it or not (this is all in Mac Lane). So I'm not sure what's particularly slick about this approach.

Anyway, the non-pointwise right Kan extension is given, for $E \overset{K}{\leftarrow} C \overset{F}{\to} D $ by $$[C,D] (G K, F) \cong [E,D] (G, \operatorname{Ran}_K F)$$ which is exactly the same as the usual definition in terms of universal arrows. The pointwise extension is $(\operatorname{Ran}_K F)e = \lim^{E(e,K-)} F$. The difference is discussed in Kelly chapter 4.3. Incidentally, Kelly sticks to pointwise extensions because the non-pointwise ones aren't much use, not because he doesn't like the usual definition.

Is that the kind of thing you're looking for?

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Yes — as Finn says, any adjunction can be re-written in your style: for a fixed $x$, saying that $\hom(Fx,y) \cong \hom(x,Gy)$ (naturally in y) is exactly the same as saying that $Fx$ represents the functor $\hom(x,G−)$. If you're happy with Yoneda-style definitions, then adjunction definitions are surely just as nice? ☺ –  Peter LeFanu Lumsdaine Jul 24 '10 at 15:23
@Finn: What's the difference between a universal arrow and a representation in enriched category theory? –  Harry Gindi Jul 24 '10 at 20:40
@Harry: see the second paragraph of Kelly ch. 1.10. –  Finn Lawler Jul 24 '10 at 21:22

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