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8 votes
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Is there a strongly noncommutative fusion category?

Consider the symmetric group group $G = S_3$ of order $6$. Then $\mathrm{H}^3_{\mathrm{gp}}(G;\mathrm{U}(1)) \cong \mathbb Z/6\mathbb Z$. Choose a generator $\omega \in \mathrm{H}^3_{\mathrm{gp}}(G;\...
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2 votes

Isomorphism between Davydov-Yetter complex and Hochschild complex of canonical algebra on a multitensor category

The category $$\mathsf{Vect}$$ behaves like a unit with respect to the Deligne tensor $$\boxtimes$$. I think the technical way to say it is that there is a canonical 2-natural equivalence $$\mathcal{...
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1 vote

TR2 for homotopy category of stable $\infty$-category

I’d like to give some more details to complement Maxime’s answer. $$\require{AMScd} \begin{CD} 1 @>>> 2 @>>> 3 \\ @VVV @VVV @VVV \\ 4 @>>> 5 @>>> 6 \\ @. @VVV @VVV \...
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2 votes

Composition map in $\infty$-categories

I had also been wondering about this for a while. I think I've figured this out. First we observe that $\operatorname{St}_{\mathcal{C}}(\{x\})=\mathfrak{C}[\mathcal{C}](-,x)$; this follows by directly ...
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4 votes
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TR2 for homotopy category of stable $\infty$-category

"This is equivalent to the assertion that the construction of the large diagram computes the suspension functor $\Sigma$" My previous answer was based on me misreading this quote :) You want ...
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10 votes
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Proposition A.2.6.15 in HTT

Retracts of weak equivalences are weak equivalences. Now if $f'$ is a retract of $f$ and you start with such a diagram with $f'$ on the left, you can create a new diagram with $f$, the same $X', X''$ ...
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0 votes

Well-behaved monad quotients

Steve Lack's paper On the monadicity of finitary monads shows that if $C$ is locally finitely presentable, then the category of finitary monads on $C$ is monadic over a power of $C$. Since monadic ...
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2 votes
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Strictness of two operations on proarrow equipments

I believe the answer to (2) is yes. First, apply the strictification theorem for bicategories twice, to make composition of arrows and proarrows both strictly associative. Thus, when our equipment is ...
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9 votes

Examples of five-adjoint systems

As noted by Simon Henry, the nLab gives many examples of adjoint chains. You can get an infinitely long adjoint chains from any ambidextrous adjunction. For an example, let $G$ be a finite group ...
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7 votes

Examples of five-adjoint systems

You can build long chains of adjoints by taking functor categories and using Kan extensions. I will give an example. Write $\underline{n}$ for the set $\{1, \ldots, n\}$ considered as a discrete ...
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7 votes

Examples of five-adjoint systems

Edit : there is actualy an nLab page listing exemple of long strings of adjunction. Maybe not what you are after, but there are examples of functors that are both left and right adjoint to each other, ...
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3 votes

Deformation of (locally) ringed spaces and of their abelian categories of modules

The answer to your second question is "no", I think. Let's assume that sufficiently nice means that it is a smooth algebraic variety over a field of characteristic $0$. Then, as written in ...
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4 votes
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Deformation of (locally) ringed spaces and of their abelian categories of modules

As Jon Pridham notes in the comments, the quote should be understood noncommutatively. In fact, in the introduction Lowen and Van den Bergh write Deformation theory of abelian categories is important ...
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5 votes
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Is every folk cofibration of strict $\omega$-categories a monomorphism?

I just thought (or maybe remember) a neat proof of this fact. It involve ideas I worked on a few years ago but never published - but that's short enough so that I can explain the key ideas on MO. Let ...
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8 votes

Why does representing functors help solving Diophantine equations?

Let me give an answer that pertains to the Diophantine equation that, according to David Speyer's answer, Lenstra was specifically talking about. How does representing functors help solve the ...
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  • 115k
6 votes

Why does representing functors help solving Diophantine equations?

Solving a Diophantine equation is the same thing as showing that the functor defined by a certain scheme $S$ of finite type over $\bf Z$ gives a non-empty set when evaluated at ${\rm Spec}({\bf Z})$. ...
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6 votes
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Naturally occurring examples of categories where composition depends on objects

Here is a way to see that Brian Shin’s example can be obtained from (standard many-hom-sets presentations of) more general constructions — so this shows clearly that constructions from the standard ...
5 votes

Naturally occurring examples of categories where composition depends on objects

How about the following. Define a category $\mathcal{C}$ with two objects $A,M$ and the following hom-sets. $\mathrm{Hom}(A,A) = \mathrm{Hom}(M,M) = \mathbb{N}$, the set of natural numbers (including ...
5 votes
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Category with domain/codomain relations

This doesn't work. In particular, we have an actual 2-isomorphism between the 2-category of these categories and the typed-definition categories: we pass back and forth simply by forming hom-classes ...
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3 votes
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Does the category of commutative and cocommutative Hopf algebras have enough injectives?

Over a field $k$, the answer is yes for injectives; I'm not sure about projectives. Over $\mathbb Z$ or other commutative rings, I really don't know -- the use of the fundamental theorem of coalgebra ...
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11 votes

Grothendieck's relative point of view and Yoneda lemma

There is a crucial aspect of the relative point of view that I think has not been completely covered here so far. The relative point of view does not just mean that we want to look at morphisms. It ...
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  • 115k
19 votes

Why does representing functors help solving Diophantine equations?

I am fairly sure the reference is to Barry Mazur's paper "An introduction to the deformation theory of Galois Representations", which is based on lectures that Mazur gave at a 1995 ...
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23 votes

Why does representing functors help solving Diophantine equations?

E.g. let $f(x,y, z)=0$ be a smooth projective plane curve with $f$ a rational polynomial of degree $\ge 4$. Then Mordell conjectured, and Faltings proved, that this has only finitely many rational ...
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4 votes

Yoneda extensions in derived categories

There is such a sequence, but it's not very interesting. Given an element of $\text{Hom}_{D^b(\mathcal{A})}(E,F[i])$, then in the same way you describe, this gives a distinguished triangle $$F\to Z_{i-...
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19 votes
Accepted

Missing axiom in the typed definition of a category?

There is no missing axiom. The notation is potentially ambiguous, but rarely (if ever) so in practice. The situation is just the same as writing addition in arbitrary abelian groups as $x + y$. ...
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1 vote

K-good trees and K-compactness of colimits over K-small downwards-closed subposets (500 point bounty if answered by Midnight EST))

Let me add a remark on a subtle point which doesn't seem to be addressed in Todd's answer. (I'm sorry I'm digging up a decade-old post!) We wanted to show that, given a $\kappa$-good $S$-tree $D:A \to ...
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  • 233
1 vote

Missing axiom in the typed definition of a category?

$\newcommand\dom{\mathit{dom}}\newcommand\codom{\mathit{codom}}\newcommand\Ar{\mathit{Ar}}\newcommand\Ob{\mathit{Ob}}\newcommand\Hom{\mathit{Hom}}\newcommand\bHom{\mathbf{Hom}}\newcommand\vertex{\...
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8 votes

Grothendieck's relative point of view and Yoneda lemma

Another statement of the relative perspective is that one should consider morphisms $X\to B$ of schemes as families of schemes (i.e. all the fibers) indexed by the base $B$. We then use this idea to ...
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  • 396
16 votes
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Grothendieck's relative point of view and Yoneda lemma

Let me answer your questions in reverse order. For the last question, yes, Yoneda's lemma is absolutely crucial to the relative point of view, as it essentially postulates that passing from a scheme $...
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9 votes
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Is there a Dold-Kan theorem for circle actions?

No, they are not equivalent, even for $C = Sp$. Indeed, the category of spectra with $S^1$-action is also the category of $\mathbb S[S^1]$-modules, and is compactly generated by a single object. On ...
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1 vote

When forgetting structure doesn't matter

For $k$ big enough, any $C^k$ map between finite-dimensional vector spaces which commutes with scalar multiplication is a homomorphism.
3 votes
Accepted

Is there a "duality involution" on presentable categories?

The answer is no, even if you restrict to the full subcategory of $Pr^L$ spanned by the $Psh(C)$'s. I'll answer in the $1$-categorical case but : a- the $\infty$-categorical case follows because ...
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5 votes

Is there a "duality involution" on presentable categories?

$\newcommand\Pr{\mathit{Pr}}\newcommand\Pres{\mathit{Pres}}$Not an answer, but too long for a comment. Gabriel–Ulmer duality lends some intuition here. For simplicity I consider the finitely ...
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5 votes
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Is $\mathit{Topos}^\text{op} \to \mathit{Pr}^L$ monadic?

Regarding monadicity (rather than comonadicity), the (2-categorical variant of the) question is answered in Bunge–Carboni's The symmetric topos. In their paper, $\mathbf A$ denotes the 2-category of ...
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  • 5,045
6 votes

Non-trivial automorphisms and descent

Briefly, descent is an analogue of taking quotients. In the category of sets, we have the following familiar facts: an equivalence relations on a set is a relation $R \subseteq A \times A$ satisfying ...
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5 votes

Non-trivial automorphisms and descent

Roughly speaking, a topos in the sense of Grothendieck is the category of sheaves on a kind of generalised space whose “points” may have non-trivial automorphisms. Question 1: What does that mean? ...
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3 votes

When forgetting structure doesn't matter

An important example from type theory/categorical logic: the forgetful functor from categories with families (or variants) to categories with attributes (or variants). Briefly, both are category-based ...
4 votes

When forgetting structure doesn't matter

An example from a pedagogical perspective: the category of based vector spaces, i.e. vector spaces with a chosen basis, and morphisms matrices, is equivalent to the category of vector spaces by the ...
2 votes

When forgetting structure doesn't matter

What about the classical theorem that the natural transformation $\eta$ witnessing the adjointess of two functions $F, G$: $$ \alpha_{X, Y} : Hom(FX, Y) \simeq Hom(X, GY) $$ is determined by the ...
4 votes

When forgetting structure doesn't matter

Group objects in the category of smooth manifolds (aka Lie groups) are the same thing as monoid objects in the category of smooth manifolds whose underlying monoid is a group. Said in other words: the ...
1 vote

When forgetting structure doesn't matter

Let $\mathcal{F}$ be the forgetful functor from the category of compact uniform spaces (the compact uniform spaces are the complete and totally bounded uniform spaces) to the category of compact ...
19 votes

When forgetting structure doesn't matter

By the positive solution to Hilbert's fifth problem (as well as the theorem of Cartan that continuous homomorphisms between Lie groups are automatically smooth), the forgetful functor from the ...
10 votes

When forgetting structure doesn't matter

The forgetful functor from abelian varieties to pointed schemes is fully faithful, (hence an equivalence onto its essential image). That is, a pointed map of schemes between abelian varieties ...
5 votes

When forgetting structure doesn't matter

There is an obstruction to nontrivial examples. With strong enough metatheory (with enough global choice), equivalent categories of structures (in which the isomorphism class of an object is a proper ...
0 votes

When forgetting structure doesn't matter

$\mathrm{C}^*$-algebras $\to$ Banach algebras that happen to admit an involution satisfying the $\mathrm{C}^*$-algebra axioms.
14 votes

When forgetting structure doesn't matter

The forgetful functor from group objects in abelian groups to abelian groups (just forget the extra group structure) is an equivalence. More generally there are lots of "idempotent" ...
8 votes

When forgetting structure doesn't matter

One reason to believe that this phenomenon is ubiquitous yet not so interesting is the following construction: Given an equivalence of categories $F \colon \mathscr C \stackrel\sim\to \mathscr D$, ...
11 votes

When forgetting structure doesn't matter

My favorite such functor is the forgetful functor from "$C^\infty$ manifolds with real coefficients" (defined as spaces with a sheaf of $\mathbb{R}$-algebras locally isomorphic to the sheaf $...
29 votes

When forgetting structure doesn't matter

The functor that takes a simplicial abelian group to its associated chain complex is arguably a "forgetful" functor. By the Dold-Kan theorem it induces an equivalence of categories.

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