25

Let $C$ be a category enriched over finite-dimensional $k$-vector spaces. A Serre functor for $C$ is a $k$-linear automorphism $S : C \to C$ such that there is a natural equivalence
$$\text{Hom}(x, y) \cong \text{Hom}(y, Sx)^{\ast}.$$
Serre functors are unique when they exist. The example that motivates the name occurs when $C = D_b(X)$ is the bounded ...

answered Jun 16 '16 at 19:20

Qiaochu Yuan

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21

Of course, Zhen has answered the question correctly under the reasonable assumption that what is being asked is whether the usual forgetful functor $U: \mathrm{Tych} \to \mathrm{Set}$ is monadic (strictly speaking, it's functors that are or are not monadic, not categories!). One might wonder whether there is some other weird functor $G: \mathrm{Tych} \to \...

answered Jun 28 '13 at 16:49

20

Broadening the question a bit, you can ask the same question about adjoint 1-morphisms in a 2-category (you're asking about 1-morphisms in the 2-category of categories). Then the 2-dimensional framed bordism 2-category gives a great example which is relatively elementary to understand. Furthermore it "explains" many of the other examples in a sense.
A ...

19

No. In fact any full subcategory of $\mathbf{Top}$ that contains all the discrete spaces cannot be monadic over $\mathbf{Set}$ unless it contains only discrete spaces. Indeed, for any such subcategory $\mathcal{C}$, the forgetful functor $\Gamma : \mathcal{C} \to \mathbf{Set}$ has a left adjoint $\Delta : \mathbf{Set} \to \mathcal{C}$ which sends a set to ...

18

Here's a really trivial way to see that the answer is "no": a functor from the empty category to a nonempty category is never a composite of adjoints (since a functor from the empty category to a nonempty category is never an adjoint).

18

In EGA 0.1.5.2-3 (from the 1971 Springer edition) the right adjoint and the left adjoint of a functor $F$ are denoted by $F^{\rm ad}$ and ${}^{\rm ad}\!F$, respectively.

15

Well, yes: the left adjoint of a functor $G: C \to D$ is the initial object in the category whose objects are pairs $(H: D \to C, \eta: 1_D \to G H)$ where $\eta$ is a natural transformation, and whose morphisms $(H, \eta) \to (H', \eta')$ are natural transformations $\theta: H \to H'$ such that
$$\begin{array}{ccc} & 1_D & \\\\ {}^{ \eta} \...

answered Nov 25 '12 at 3:37

15

Let me start by discussing a bit the option of having a large class of generators. You might be interested in the notion of locally class-presentable.
To be precise here, I need to be a bit set-theoretical, thus, let me start with an informal comment.
Informal comment. Indeed your category is locally class-presentable, class-accessibility is a very ...

14

Let $\mathcal C$ and $\mathcal D$ be two categories, and let $F\colon\mathcal C\longrightarrow \mathcal D$ and $G\colon\mathcal D\longrightarrow\mathcal C$ be two functors, with $F$ left adjoint to $G$. Then there are natural transformations $F\circ G\longrightarrow \operatorname{Id}_{\mathcal D}$ and $\operatorname{Id}_{\mathcal C}\longrightarrow G\circ F$ ...

answered Dec 1 '18 at 18:54

Leonid Positselski

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12

The answer is no. Let $C$ be a category such that the unique map from $C$ to the terminal category is a composition of $n$ adjoints. Then $C$ has an object $x_0$ such that every other object of $C$ can be connected to $x_0$ by a zigzag of length at most $n$; this is easy to prove by induction on $n$.
In particular, let $R$ be the "infinite zigzag", the ...

12

Let $C$ and $D$ both be the category of finite-dimensional (say real) vector spaces and invertible linear maps between them, let $F$ be the identity, and let $G$ take a vector space to its dual. ($G$ is not functorial on all linear maps, but it is on the invertible ones.) Then $G y \cong y$ unnaturally, so $\hom(F x, y) = \hom(x, y) \cong \hom(x, G y)$ ...

11

The empty category trivially satisfies this (there are no functors at all from a nonempty category to the empty category), but no other such category exists. Let $A$ be any category with a terminal object $1$, and consider the projection $C\times A \to C$. This has a right adjoint $C\to C\times A$ given by $c\mapsto (c,1)$. However, these functors are not ...

11

Another and probably more natural interpretation of the sentence in the Wikipedia article may be called "localizing an adjunction to an equivalence".
Let $\mathcal C$ and $\mathcal D$ be two categories, and let $F\colon\mathcal C\longrightarrow \mathcal D$ and $G\colon\mathcal D\longrightarrow\mathcal C$ be two functors, with $F$ left adjoint to $G$. Then ...

answered Dec 1 '18 at 22:51

Leonid Positselski

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11

Let $\mathcal{C},\mathcal{D}$ two $\infty$-categories and $f:\mathcal{C}\to\mathcal{D}$ and $g:\mathcal{D}\to \mathcal{C}$ two functors.
Recall (HTT.5.2.2.7) that a natural transformation $u:1_{\mathcal{C}}\to gf$ is a unit transformation if the natural transformation
$$(\ast)\qquad\mathrm{Map}_{\mathcal{D}}(f-,-)\xrightarrow{g} \mathrm{Map}_{\mathcal{C}}(...

10

Probably you had some trouble finding this because the search term $2$-$\text{Cat}$ is not accurate enough; you want not the cartesian monoidal product on $2$-$\text{Cat}$, but rather what is called the Gray monoidal product; the tricategory you want then is denoted $\text{Gray}$, the tricategory of strict 2-categories, strict 2-functors, pseudonatural ...

answered Mar 2 '15 at 14:02

10

For the first question: Yes.
Let $C$ be a category with finite products and let $\Delta=(\Delta_1,\Delta_2):C\to C\times C$ s.t. $$\times\dashv \Delta.$$
More specifically we find $$[X\times Y, Z]\simeq[(X,Y),\Delta Z]=[X,\Delta_1 Z]\times[Y,\Delta_2 Z].$$ Note that this isomorphism is functorial in all arguments and given $f:X'\to X$ and $g:Y'\to Y$ we ...

10

Wow, I have always thought that unstraightening has to be easier than straightening, but I've never actually looked at Lurie's treatment before, so I'm surprised to realize he defines straightening directly while defining unstraightening as its right adjoint.
Anyway, I'm pretty sure that Remark 2.2.2.11 follows from the description of straightening over a ...

9

In case $X, Y$ are smooth,
$ f^* $
is full if and only if it is full and faithful. This is explained in the introduction of arXiv:1101.5931 (by Canonaco-Orlov-Stellari), which also studies when this implication holds more generally. Thus the pull-back is full and faithful if and only if $Rf_* \mathcal{O}_X = \mathcal{O}_Y$; for example, when $f$ is ...

9

Another widely used example of infinite adjunction chains arises in linguistics, specifically in connection with pregroup grammars. It has been first observed by Lambek in Some Galois Connections in Elementary Number Theory (J. Number Theory 47 (1994), 371-377). Take monotone maps $f:\mathbb Z\to\mathbb Z$ unbounded in both directions (monotone with respect ...

answered Jun 17 '16 at 22:19

მამუკა ჯიბლაძე

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9

This is a well-known result in category theory: that any equivalence can be improved to an adjoint equivalence. Given an equivalence in the form of isomorphisms $\eta: 1_C \stackrel{\sim}\to GF$ and $\xi: FG \stackrel{\sim}\to 1_D$, what we would like is for $\eta$ to be the unit and $\xi$ the counit of an adjunction, and for this we need the famous ...

answered Nov 3 '16 at 4:58

9

Building on Garlef Wegart's work, the answer is yes in general. The idea is to show that if $C$ has a right adjoint to $\times$, then it is enriched in pointed sets, so that $C$ embeds fully faithfully in a category $C_\ast$ with a zero object and the easy argument can be applied. And as noted already, in this case $C$ will have biproducts with respect to a (...

8

Your question is not as precise as you portray it, and apart from the question of adjointness, a naive
treatment of definability like this leads easily to
contradictions. Specifically, I claim that your definitions do not
actually succeed in defining the concepts of definability and of
being a set-builder.
To convince you of this, let me prove that the ...

answered Oct 29 '13 at 23:29

Joel David Hamkins

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8

I really like this question, I've been trying to sort out some of these ideas for a little while. I don't know the answer to your questions about conilpotence and twisting morphisms vs twisted arrows. I do have reason to believe that twisted arrows between A and C are the same as the twisted arrows from A to conil(C) but I don't know how to prove that.
I ...

8

Another nice example is the infinite sequence of adjunctions characterizing stable homotopy theories. One has, in any homotopy theory $K$ (whatever you think that is, as long as there is a notion of (homotopy) limits and colimits) a sequence $1^*\vdash \pi^*\vdash 0^*$ of adjunctions between $K$ and the homotopy theory $K^{[1]}$ of morphisms in $K$, where's $...

8

I understand it like this: if a monad $M: C \to C$ has a right adjoint $K: C \to C$, then that right adjoint carries a comonad structure which is mated to the monad structure, and the category of $M$-algebras is canonically equivalent to the category of $K$-coalgebras. All examples of functors that are simultaneously monadic and comonadic are of this type. ...

answered Jul 17 '17 at 0:16

8

I don't know what morphisms you intend for the category of finite-dimensional Hilbert spaces, but it doesn't actually matter. The answer is no, there are no interesting adjunctions between the category of groupoids and the category of finite-dimensional Hilbert spaces.
More generally,
If $C$ is a complete and cocomplete category and $D$ is a small ...

7

It's a good question; the answer is no.
Suppose $F \dashv F^{op}$ with unit $\eta: 1_C \to F^{op} F$ and counit $\varepsilon: F F^{op} \to 1_{C^{op}}$. Since $\varepsilon$ is the unique transformation $\theta: F F^{op} \to 1$ such that
$$1_{F^{op}} = (F^{op} \stackrel{\eta F^{op}}{\to} F^{op} F F^{op} \stackrel{F^{op}\theta}{\to} F^{op})$$
the question ...

answered Sep 3 '12 at 12:42

7

Given any pair of sets $X, Y$ (let me not be too specific about what "sets" means because what follows is robust with respect to changes in the definition), any relation $R : X \times Y \to 2$ whatsoever induces a contravariant adjunction between the poset of subsets of $X$ and the poset of subsets of $Y$ given by
$$2^X \ni S \mapsto F(S) = \{ y \in Y : xRy ...

answered Oct 29 '13 at 17:19

Qiaochu Yuan

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7

let me offer you an alternative. The free 2-group on a groupoid is well defined up to equivalence (not isomorphism), hence I'll offer you a strict model (which is known to exist by abstract reasons, since any 2-group can be strictified).
A strict 2-group is essentially the same thing as a crossed module $\partial\colon\mathcal F(G)_1\rightarrow\mathcal F(G)...

7

While Paul's answer is correct, it is also possible to start with an ordinary category $C$ and "enrich it" by adding structure stuff. One sort of tautological way to do this is to specify a $V$-enriched category $\hat{C}$ whose underlying ordinary category (in the sense mentioned by Todd) is isomorphic (or equivalent) to $C$. A more interesting way is to ...

answered Dec 30 '15 at 6:36

Mike Shulman

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