## Hot answers tagged adjoint-functors

56

The adjoint functor theorem as stated here and the special adjoint functor theorem (which can also both be found in Mac Lane) are both very handy for showing the existence of adjoint functors.
First here is the statement of the special adjoint functor theorem:
Theorem Let $G\colon D\to C$ be a functor and suppose that the following conditions are ...

45

Lots of people-who-are-fond-of-adjoint-functor-theorems have responded to this post saying "adjoint functor theorems". Let me give a more mundane and rather different answer which fits much better into my world view.
In my experience (which may differ from others), the true answer is that category theorists have these adjoint functor theorems which work ...

39

The answer is no, because the nerve functor turns an adjoint pair of functors between categories into inverse homotopy equivalences between spaces (this is because of the existence of the unit and counit and the fact that nerve turns natural transformations into homotopies). In particular, this means that any functor whose nerve is not a homotopy ...

37

Other people have mentioned the Adjoint Functor Theorems. Here's a different perspective.
There's a famous Cambridge exam question set by Peter Johnstone:
Write an essay on
(a) the usefulness, or
(b) the uselessness,
of the Adjoint Functor Theorems.
I agree with the undertone of the question: the Adjoint Functor Theorems (AFTs) aren't as ...

28

The example I would give a five-year old is the following. Take the category R whose objects are real numbers (or perhaps rational numbers for the five-year old) and a single
morphism x--> y whenever x is less than or equal to y. Let Z be the full subcategory consisting of the integers. The inclusion i:Z --> R has a left and a right adjoint: one is the ...

22

There cannot be a "free coalgebra" functor, at least in what I think is the standard usage. Namely, suppose that "orange" is a type of algebraic object, for which there is a natural "forgetful" functor from "orange" objects to "blue" objects. Then the "free orange" functor from Blue to Orange is the left adjoint, if it exists, to the forgetful map from ...

21

I like Wikipedia's motivation for an adjoint functor as a formulaic solution to an optimization problem (though I'm biased, because I helped write it). In short, "adjoint" means most efficient and "functor" means formulaic solution.
Here's a digest version of the discussion to make this more precise:
An adjoint functor is a way of giving the most ...

17

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 ...

17

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 ...

17

Yes, there are many such functors. They are usually called "biadjoint." A good example is tensor product with a vector space $V$ in the category of finite dimensional vector spaces. This is actually adjoint to itself.
This is a little funny since to find this adjunction you have to pick an isomorphism $V\cong V^*$, but that's OK; adjunction of functors ...

16

There are lots of examples. Here's what I think is in some sense the minimal one.
Let $C$ be the terminal category $\mathbf{1}$ (one object, and only the identity arrow). Then for any category $D$, a left adjoint to the unique functor $G: D \to \mathbf{1}$ is an initial object of $D$, and a right adjoint is a terminal object. So, we're looking for a ...

15

One way to understand $l^1(X)$ for a set $X$ with counting measure is that $l^1(-): Set \to Ban$ provides a left adjoint to the functor $\hom(k, -): Ban \to Set$. Here $k$ is the ground field and $Ban$ denotes the category of Banach spaces and linear maps $T: X \to Y$ with $\|T\| \leq 1$. Another way of saying this is that $l^1(X)$ is a coproduct of an ...

13

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).

13

Obligatory n-lab reference: adjoint functor theorems
Figuring out when functors had adjoints or not was something I did a lot of in Comparative Smootheology (section 8).
Edit: Thought I'd expand on my comment to Andrew Critch's answer. A simple application of the Special Adjoint Functor Theorem is to universal algebra where it becomes:
Theorem Let $D$ ...

13

I think that it is highly unlikely that there exists a separification functor. What does exist is the following:
Theorem (Raynaud-Gruson): Let S be a base scheme and work relative to S. Given a non-separated scheme X of finite type, there is a blow-up (a proper birational morphism) X'->X such that X' admits an étale morphism to a projective scheme Z (in ...

13

The article here proves that the forgetful functor from $k$-Hopf algebras to $k$-algebras has a right adjoint. The main tool they use is the Special Adjoint Functor Theorem.

12

The following is really an adjunction between $2$-categories but I am going to ignore that subtlety. This blog post discusses everything in more detail and with a few more examples.
Consider on the one hand all concrete categories, by which I mean pairs $(C, U)$ of a category $C$ and a functor $U : C \to \text{Set}$ (the "underlying set") functor, and on ...

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

Theo, your question is in the neighborhood of what is called "total cocompleteness" or "totality". A category $C$ is total if the Yoneda embedding $y: C \to Set^{C^{op}}$ has a left adjoint. There is a similar notion of totality in the enriched case.
Total categories have this "saft" property you are discussing: every cocontinuous functor from a total ...

11

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} ...

11

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 ...

11

I think frequently the easiest to check sufficient conditions are the following, which also make precise why "this looks like it might work" is so often successful:
Theorem: A functor $G: C\to D$ is a right adjoint functor (i.e. has a left adjoint) if and only if for each object $Y$ in $D$, there exists an initial morphism $\phi_Y:Y\to G(I_Y)$ from $Y$ to ...

11

I do not know the answer for your question, but let me rephrase the warning that takes places right after Definition 4.17 (p.21) of certain unrelated lecture notes:
Contrary to general belief, the coalgebra $T(V)$ with the projection $T(V) → V$
is not cofree in the category of coassociative coalgebras! Cofree coalgebras (in the sense of
the obvious ...

10

There are two parts to this answer.
First, a functor must be continuous (cocontinuous) to have a left (right) adjoint. Most of the times, it is easy to check that a functor does not preserve (co)limits and thus it cannot have a a left (right) adjoint.
(co)continuity is not enough to actually prove that a functor has the required adjoint, but it is almost ...

10

As I understand things, the desire to find a more categorical construction of the free abelian group on a set S is not itself fully in the categorical spirit. Rather, for an object which is defined by a universal mapping property, you only need to convince yourself that it exists; you don't need to be bothered by looking at any particular construction. The ...

10

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 ...

10

Let CBA denote the category of commutative, unital Banach algebras, with morphisms being the continuous unit-preserving homomorphisms. Let CHff denote the category of compact Hausdorff spaces, with morphisms being the continuous maps.
There's a contravariant functor $C$ from CHff to CBA (or if you prefer, a functor $C$ from CHff${}^{\rm op}$ to CBA) which ...

9

The usual functor $L:\text{LieGroups}\to\text{LieAlgebras}$ has a left adjoint $\Gamma$. The unit $A\to L\Gamma A$ is an isomorphism, but the counit $\Gamma L G\to G$ is the universal covering of the base component of $G$.
The functor $\pi_0$ is left adjoint to the inclusion of discrete groups in Lie groups.
The category of tori is equivalent to the ...

9

The answer of Ben Webster, can be made easier. Consider the functor F : (A-mod) -> (A-mod) which maps any A-module on (0). Then, F is a left adjoint to F ; and so, is a also a right adjoint to F. This is clear because for all A-modules N, M, one has Hom_A(0,N)=Hom_A(M,0). But, F is not an equivalence.

9

Always.
In the case of algebraic theories, the proof is simple. Let $F : Set \to V$ be the free functor, where $V$ is some algebraic theory. Then if $X$ is a set and $U$ a $V$-algebra,
$$
\operatorname{Hom}_{V}(F(X),U) \cong \operatorname{Hom}_{\operatorname{Set}}(X,|U|)
$$
The right-hand side is a $V$-algebra, naturally in both $X$ and $U$. ...

Only top voted, non community-wiki answers of a minimum length are eligible