14 votes

Conjugacy classes as left Kan extension of forgetful functor

$\newcommand{\Gp}{\mathbf{Grp}} \newcommand{\conj}{^{\mathbf{conj}}} \newcommand{\Gpd}{\mathbf{Gpd}} \newcommand{\Set}{\mathbf{Set}} \newcommand{\ho}{\mathrm{ho}} \newcommand{\Fun}{\mathrm{Fun}} \...
Maxime Ramzi's user avatar
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7 votes
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The Kan construction, profunctors, and Kan extensions

I will try to answer the second question. Prop 1. Let ${\bf C} \xleftarrow{i} {\bf A} \xrightarrow{f} {\bf B}$ be a span where $i$ is dense and fully faithful. Moreover $\text{lan}_if$ is ...
Ivan Di Liberti's user avatar
7 votes
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Is there such a thing as a weighted Kan extension?

Yes. Given $F:C\to D$ and a profunctor $H:E$ ⇸ $C$, i.e. a functor $H : C^{\rm op}\times E\to \rm Set$ (or to the enriching category $V$), the $H$-weighted colimit of $F$ is the functor $L : E \...
Mike Shulman's user avatar
7 votes
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How to understand adjoint functors?

Nice question Bumblebee. So, let us start with some "metaphysics of adjointness": THE LEFT AND RIGHT ADJOINTS TO A FUNCTOR $ \mathcal{F}:\mathcal{C}\hookrightarrow\mathcal{D}$ ARE THE FREE (...
Mirco A. Mannucci's user avatar
7 votes

Conjugacy classes as left Kan extension of forgetful functor

An alternative to Maxime's sophisticated proof is to observe that $X$ and $L$ are both corepresented by the free group, because a group homomorphism from a free group (up to conjugacy) is the same as ...
Shay Ben Moshe's user avatar
6 votes
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Base change isomorphism for left Kan extensions

I believe one set of conditions is for either $\varphi$ to be proper or $\psi$ to be smooth. The dual of this (using right Kan extensions rather than left Kan extensions) is proven by Cisinski in &...
daniel gratzer's user avatar
6 votes
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When Kan extensions don't exist

Yes, directions like this have been explored, for all kinds of objects with universal properties (which includes Kan extensions, since as MacLane famously wrote "all concepts are Kan extensions"). ...
Mike Shulman's user avatar
6 votes

Existence of pointwise Kan extensions in $\infty$-categories

Lurie's approach First let me explain why this is already in HTT 4.3.3. Recall that a (pointwise) left Kan extension of $F: A \to \mathcal{C}$ along an inclusion $i: A \to B$ is a functor $F: B \to \...
Dylan Wilson's user avatar
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6 votes
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Kan extensions inside a monoidal category

It is certainly the case that the duals of internal homs have appeared significantly less in the categorical literature. I've included a few more references below, but I am not sure this is a ...
varkor's user avatar
  • 8,675
6 votes

Is the singular simplicial functor full

Any non-discrete totally disconnected space gives a counterexample, e.g. the Cantor space $2^{\mathbb{N}}$. If a space $X$ is totally disconnected, then every simplex in it is constant (since the ...
Peter LeFanu Lumsdaine's user avatar
5 votes
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Yoneda extension preserving finite products?

(Update: the following answer applies only in the case that $C$ has finite products.) The left Kan extension $\hat{F}$ preserves finite products just when $F$ does. One direction is easy since $F \...
john's user avatar
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4 votes
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Left and right Kan extensions

That is indeed the right answer: the left adjoint is determined by what it does on (the Yoneda image of) $\mathcal{C}$, since it is a colimit-preserving functor, and there we have $$\mathrm{Hom}_{\...
Rune Haugseng's user avatar
4 votes

Kan extensions in the $2$-category of monoidal categories

Here are some miscellaneous remarks/thoughts on these notions. Motivation for Monoidal Co/Limits and Variant Notions. (You've said part of this already, but let me start here for context) If we want ...
Emily's user avatar
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4 votes

How to understand adjoint functors?

If you start with a category and only consider what you can see by looking at functors from groupoids, well you’ll only see the invertible morphisms. So the right adjoint is the core. If you start ...
Noah Snyder's user avatar
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4 votes
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Question about the proof of Kerodon tag 030V (Proposition 7.3.7.1)

I think you are right that the cube is not a levelwise equivalence - I saw you already put a comment on the Kerodon page so you should get an answer soon. Here's how I would fix the proof: call your ...
Maxime Ramzi's user avatar
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4 votes

Adjoining extensions in bicategories

A partial answer is contained in Betti's Formal theory of internal categories (page 49), where he states that the bicategory $\mathbf{Dist}(\mathcal E)$ of $\mathcal E$-internal distributors is the ...
varkor's user avatar
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3 votes
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Relationship between Kan extensions and internal hom

The analogous adjunction for the enriched category of functors can be deduced from the adjunction for the ordinary category of functors using the Yoneda lemma. Indeed, to establish a natural ...
Dmitri Pavlov's user avatar
3 votes

When Kan extensions don't exist

There are several points worth making here. (i) One solution is to use Benabou's theory of 'distributeurs' also called profunctors. A functor F from C to D defines two profunctors basically $D(F-,-):...
Tim Porter's user avatar
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3 votes

Do (co)density (co)monadic constructions stablize?

Not sure about state of the art, but here are a couple of comments. As is pointed out in your reference, if $F$ has a left adjoint $G$, then $T=FG$. If $F$ happens to be fully faithful, then $G$ is a ...
Gregory Arone's user avatar
3 votes

Kan extensions in the $2$-category of monoidal categories

I am not an expert, but I think the best reference for this topic is Algebraic Kan extensions along morphisms of internal algebra classifiers by Mark Weber.
Ivan Di Liberti's user avatar
3 votes
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Faithfulness of Right adjoint to Kan extension

Assuming $C$ is a small category, the right adjoint to $Lan_y(F): PShv(C) = Set^{C^{op}} \to D$ is just the functor $D \to Set^{C^{op}}: d \mapsto \hom_D(F-, d)$; this basic "generalized nerve" ...
Todd Trimble's user avatar
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3 votes
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Existence and characterisations of left Kan extensions and liftings in the bicategory of relations I

Here is a partial answer, addressing (3). Given a relation $S : A⇸B$, composition with $S$ is a cocontinuous order-preserving map $Rel(B,C) \to Rel(A,C)$. These hom-posets are powerset lattices $2^{B\...
Tim Campion's user avatar
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2 votes

Kan extension of conservative functors

No, take $A$ to be any cocomplete category. For any category $I$ the right Kan extension of $Id:A\to A$ along the embedding $\iota:A\to Fun(I,A)$(object $a$ goes to the constant functor $\underline{a}$...
SashaP's user avatar
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2 votes
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About pointwise Kan extension

The answer is no (I think -- non-pointwise Kan extensions are a pain and I may have messed something up!). I wouldn't lose too much sleep over this, though -- in practice, you never know that some ...
Tim Campion's user avatar
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2 votes

Are left and right Kan extensions ever isomorphic?

Sure. Consider the left and right Kan extension along the terminal object $t: 1 \to \text{Set}$, applied to a functor $X: 1 \to \text{Sup}$ in the category of sup-lattices. The left Kan extension $\...
Todd Trimble's user avatar
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2 votes
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(Pro-)representable functors and full subcategories in homotopy theory

This is a partial answer. Broadly speaking, representability theorems break down into two types. In both cases, the functor $F$ has to satisfy some exactness condition. For Freyd type theorems, $F$ ...
David White's user avatar
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2 votes
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Weighted limits and Kan extension in Dist

I'm not sure I've followed your question exactly, so let me rephrase it to how I've understood it, and you can tell me if I've misunderstood. I shall use different letters to make sure I'm not ...
varkor's user avatar
  • 8,675
1 vote

Pointwise Kan extensions VS weighted limits

I think the equivalence becomes trivial if you add the hidden assumption, that is the hypothesis of representability of the extension in $Dist$ This is akin to Kelly's remark about limit in enriched ...
nicolas's user avatar
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1 vote

Kan extensions inside a monoidal category

The common phrase is coexponential object. The internal logic of coclosed categories is cointuitionistic logic. You can broadly think of cointuitionistic logic in terms of pattern matching, ...
Molly Stewart-Gallus's user avatar
1 vote

Extending monads along dense functors

This is only a partial answer to your questions, but one that I imagine recovers all the cases of interest (in particular it gives a conceptual reason why your examples hold when $j$ is the Yoneda ...
varkor's user avatar
  • 8,675

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