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Recall that a Kan extension is called pointwise if it can be computed by the usual (co)limit formula, or equivalently if it is preserved by (co)representable functors.

I have seen pointwise Kan extensions defined in many texts, such as Mac Lane's book or Borceux's book, but these texts don't typically prove any theorems about them. I've also heard it claimed that all mathematically important Kan extensions are pointwise. And apparently it should be easier to compute a pointwise Kan extension since there is a formula for it. But I'd like to know: what are some interesting categorical facts that are true about pointwise Kan extensions that are not true about general Kan extensions? I'm happy with properties whose proofs don't generalize to the non-pointwise case, but I'd also be interested to see actual counterexamples to such generalizations.

In Basic Concepts of Enriched Category Theory, Kelly chooses to reserve the term "Kan extension" for the pointwise version. In explaining this choice, he says "Our present choice of nomenclature is based on our failure to find a single instance where a weak Kan extension plays any mathematical role whatsoever." If I comb through his book for uses of Kan extensions, will I find that "most" facts used about Kan extensions fail in the non-pointwise case?

One fact which I have seen stated is that in a Kan extension $\require{AMScd}$ \begin{CD} A @>G>> C\\ @VFVV \nearrow \mathrm{Lan}_F G \\ B \end{CD}

If $G$ is fully faithful and $\mathrm{Lan}_F G$ is pointwise, then the comparison 2-cell is an isomorphism. Is this false when the pointwiseness hypothesis is dropped?

Part of the reason I ask is that there are several definitions in the literature generalizing pointwiseness of Kan extensions to more general categorical frameworks (enriched categories, 2-categories, equipments, double categories...) but I'm confused because I don't know what the theory is that these theories are attempting to generalize. Pointwiseness also shows up in the theory of exact squares, but again I'm confused because I don't know the point of pointwise Kan extensions.

EDIT I'm starting to compile a list of these properties. Here's what I have so far:

  1. As above, a pointwise extension along a fully faithful functor has an isomorphism for a comparison cell.

  2. A functor $F$ is dense if and only if its left Kan extension along itself $\mathrm{Lan}_F F$ exists, is pointwise, and is isomorphic to the identity functor.

  3. To show that taking total derived functors is functorial, one needs that they are pointwise extensions.

  4. To show that the total derived functors of a Quillen adjunction form an adjunction between homotopy categories, one needs to know that they are pointwise extensions.

(3) and (4) were mentioned by Emily Riehl in introductory remarks to a lecture at the Young Topologists' Meeting 2015 (link).

  1. Pointwise Kan extension along a fully faithful functor is itself a fully faithful functor. More precisely, if $F: A \to B$ is fully faithful, and $\mathrm{Lan}_F G$ and $\mathrm{Lan}_F G'$ exist and and are pointwise, then $\mathrm{Nat}(G,G') \to \mathrm{Nat}(\mathrm{Lan}_F G, \mathrm{Lan}_F G')$ is an isomorphism.

So far the only counterexamples I have to dropping the pointwise hypothesis in these statements are in the cases (1) and (2), which I've detailed in CW responses.

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    $\begingroup$ "I've also heard it claimed that all mathematically important Kan extensions are pointwise." [Citation Needed] $\endgroup$ – Vidit Nanda Oct 7 '15 at 3:21
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    $\begingroup$ circular comment: isn't the point of pointwise Kan extensions precisely the fact that they are pointwise? I mean, the colimit formula is for me really useful. Not sure if I can pinpoint a precise example, but when I was learning some basics of "functorial" algebraic geometry, having a formula to compute Kan extensions was important (both for understanding and computing). $\endgroup$ – pro Oct 7 '15 at 3:26
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    $\begingroup$ @ViditNanda :) I added a paragraph with the Kelly quote I was thinking of. One of the more surprising instances of this phenomenon (emphasized by Emily Riehl in Categorical Homotopy Theory) is the fact that derived functors, when computed via functorial resolutions, are not only pointwise, but absolute Kan extensions (preserved by any functor whatsoever) -- so they can be computed via the (co)limit formula even though derived categories don't have many (co)limits. $\endgroup$ – Tim Campion Oct 7 '15 at 3:28
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    $\begingroup$ @TimCampion silly example from DAG: let's work over the complex numbers, if R is a cdga (say in negative degrees) there is an $\infty$-cat of R-modules. Say we are interested only in the maximal groupoid (because we want to do stacks) and call it R-Mod (it's an $\infty$-groupoid, ie a space). The construction extends to a whole functor cdga --> Spaces. Now, say you have an arbitrary prestack F on cdga's (ie a functor cdga --> Spaces), how do you define the category of quasi-coherent modules over it? Well, it's a Kan extension of the previous functor along the inclusion cdga--> Fun(cdga,Spaces) $\endgroup$ – pro Oct 7 '15 at 13:32
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    $\begingroup$ I guess I didn't really need to go full derived. One can run the same story for ordinary rings R and the abelian category R-Mod. The issue is that for an arbitrary presheaf F, that construction does not yield an abelian category in general (while the derived counterpart spits out a stable category, which is what you want). $\endgroup$ – pro Oct 7 '15 at 13:37
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I have always thought that pointwise Kan extensions are better than normal Kan extensions because you can actually compute them using a (co)end

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    $\begingroup$ Strictly speaking the (co)end formula only applies if the codomain category is (co)powered over the enriching category, although in most cases it will be. But anyway, here's what should be an easy question: what's an example of something you can do with such a computation that you couldn't do with just the universal property of a Kan extension? $\endgroup$ – Tim Campion Oct 17 '15 at 14:34
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    $\begingroup$ Let $f : A \to B$ be a map of rings and $M$ an $A$-module. Then $B \otimes_A M$ is a pointwise left Kan extension computed as a coend. The universal property of tensor products is useful, but often it is useful to manipulate elements in $B \otimes_A M$ directly. More generally, if you repace $B$ and $A$ with ringoids (something that I do frequently) it is still useful to do explicit computations instead of just relying on the universal properties. $\endgroup$ – Daniel Barter Oct 17 '15 at 14:41
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    $\begingroup$ Cool! So this is in the additive context. Now, I'm probably being perverse, but let me press the issue as I did with pro in the comments above. When do you need to use the elements of $B \otimes_A M$ directly? 90% of the time, I think one simply maps out on elements of the form $b \otimes m$, which I think can equivalently be done using the universal property of the Kan extension. $\endgroup$ – Tim Campion Oct 17 '15 at 14:50
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    $\begingroup$ One situation where you need to use the elements directly is when you map into a tensor product. This happens a lot when you think about Morita theory. Often the indecomposable projectives that you are working with are induced up from some smaller ring(oid) and you need to compute the maps between these indecomposable projectives $\endgroup$ – Daniel Barter Oct 17 '15 at 14:54
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    $\begingroup$ Hm. So I guess this reduces the question to something about the connection between computability in category theory and (co)limits: (co)limits, when they exist, tend to be computable, while pretty much anything that's computable is some kind of (co)limit. This is particularly mysterious when you consider total derived functors: these are pointwise Kan extensions, "computable" in terms of (co)limits, even though general (co)limits are not readily computable in derived categories. $\endgroup$ – Tim Campion Oct 17 '15 at 15:18
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A functor $F: A \to B$ is dense if and only if its left Kan extension along itself exists, is pointwise, and is isomorphic to the identity. This is because the defining universal property of a pointwise extension is

$B(\mathrm{Lan}_F F (b), b') \cong [A^\mathrm{op}, \mathsf{Set}](B(F-,b),B(F-,b'))$

To say that $\mathrm{Lan}_F F$ exists, is pointwise, and is isomorphic to the identity is to say that the left hand side is $B(b,b')$, while to say that $F$ is dense is to say that the right hand side is $B(b,b')$.

Here's an example of a functor $F$ which is not dense, but such that $\mathrm{Lan}_F F$ exists and is isomorphic to the identity (in particular, the extension is not pointwise). Take $A=B$ to be the monoid of natural numbers under addition, viewed as a 1-object category $\mathbf{B}\mathbb{N}$. Let $F: \mathbf{B}\mathbb{N} \to \mathbf{B}\mathbb{N}$ be the "multiply by 2" functor. Then $F$ is not dense, but $\mathrm{Lan}_F F$ exists and is the identity $\mathrm{Id}$. For, any functor $H: \mathbf{B}\mathbb{N} \to \mathbf{B}\mathbb{N}$ is given by multiplication by some $m$, and there is a natural transformation from $\mathrm{Id} \implies H$ if and only if $H= \mathrm{Id}$. Whereas a natural transformation $F \implies HF$ exists if and only if $2n = H(2n)$ for all $n$, which implies that $H = \mathrm{Id}$. In both cases, if a natural transformation exists, then any $\alpha \in \mathbb{N}$ is natural.

The argument that $F$ is not dense goes as follows: $\mathbf{B}\mathbb{N}(F-,\bullet)$ is the set $\mathbb{N}$ but the $\mathbb{N}$-action is addition by 2 rather than 1. The image of $\mathbf{B}\mathbb{N}$ under $\mathbf{B}{N}(F-,1)$ includes all addition-by-$n$ actions on this $\mathbb{N}$-set, but there are more $\mathbb{N}$-set endomorphisms: namely, the odd numbers and even numbers can be shifted independently.

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In a Kan extension $\mathrm{Lan}_F G$, $\require{AMScd}$ \begin{CD} A @>G>> C\\ @VFVV \nearrow \mathrm{Lan}_F G \\ B \end{CD}

If $F$ is fully faithful and the extension is pointwise, then the comparison 2-cell is an isomorphism. This is because, if the extension is pointwise, we have

$C(\mathrm{Lan}_G F(Fa),c) \cong [A^\mathrm{op},\mathsf{Set}]( B(F-,Fa),C(G-,c))$

$\qquad \qquad \qquad \, \cong [A^\mathrm{op},\mathsf{Set}]( A(-,a),C(G-,c))$

$\qquad \qquad \qquad \, \cong C(Ga,c)$

and we can take $c = Fa'$.

In order to get an example of a fully faithful $F$ such that $\mathrm{Lan}_F G$ does not have an isomorphism for a comparison cell, we need to move beyond monoids, because a fully faithful functor between monoids is an isomorphism. So the simplest sort of fully faithful functor which is not an equivalence would be the inclusion $F: \mathbf{B}M \to \mathbf{B}M_+$ of a monoid into the category which freely adjoins either an initial or terminal object to it. If we adjoin a terminal object, then left Kan extensions will all be pointwise (with the terminal object being sent to the colimit of the original diagram), so we adjoin an initial object instead. And we might as well take $M$ to be the simplest possible monoid $\mathbb{N}$:

$\require{AMScd}$ \begin{CD} \mathbf{B}\mathbb{N} @>G>> C\\ @VFVV \nearrow L \\ \mathbf{B}\mathbb{N}_+ \end{CD}

It turns out there is a "minimal" $C$ admitting such a left Kan extension $L$, which looks like this:

$G\bullet \overset{(\eta_n)_{n \in \mathbb{N}}}{\overset{\to}{\to}} L\bullet \overset{L!}{\leftarrow} L\emptyset$

Here $G\bullet$ is the fully faithful image of $G$ (so it's a copy of $\mathbf{B}\mathbb{N}$), and $L\emptyset \overset{L!}{\to} L\bullet$ is the fully faithful image of $L$ (so it's a copy of $\mathbf{B}\mathbb{N}_+$ where $L\emptyset$ is the initial object). $C(G\bullet, L\bullet)$ is generated by an arrow $\eta = \eta_0$ under the equation $\eta_n = \eta \circ Gn = Ln \circ \eta$. So $\eta$ constitutes a comparison natural transformation $G \implies LF$.

An exhaustive analysis of the functors $H: \mathbf{B}\mathbb{N}_+ \to C$ (there are 5 families of them, depending on where the objects are sent) reveals that the only one which admits a natural transformation $G \implies HF$ or $L \implies H$ is $L$ itself, and this diagram is in fact a left Kan extension $L = \mathrm{Lan}_F G$ with $F$ fully faithful, but the comparison 2-cell $\eta$ is not invertible.

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