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}}
\...
7
votes
Accepted
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 ...
7
votes
Accepted
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 \...
7
votes
Accepted
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 (...
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 ...
6
votes
Accepted
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 &...
6
votes
Accepted
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"). ...
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 \...
6
votes
Accepted
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 ...
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 ...
5
votes
Accepted
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 \...
4
votes
Accepted
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}_{\...
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 ...
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 ...
4
votes
Accepted
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 ...
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 ...
3
votes
Accepted
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 ...
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-,-):...
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 ...
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.
3
votes
Accepted
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" ...
3
votes
Accepted
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\...
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 $\underline{a}$...
2
votes
Accepted
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 ...
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 $\...
2
votes
Accepted
(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, $F$ ...
2
votes
Accepted
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 ...
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 ...
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, ...
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 ...
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