A bestiary about adjunctions

Question

What is your favourite adjoint? Following Mac Lane philosophy adjoints are everywhere, so I would like to draw a (possibly but unprobably) exhaustive list of adjunctions one faces in studying Mathematics. For the sake of clarity I would like you to follow a general scheme, a very naive example of which can be the following:

1. Functors F and G between cats C and D
2. Is the adjunction a (co)reflection?
3. Does the left adjoint admit a left adjoint on its own?
4. Anything you want to add

Obviously you are totally free to expand it, revert it...

I would also like to grasp something more than a mere enumeration: i.e. listing all adjunctions $\mathbf{Groups}\leftrightarrows\mathbf{Sets}$, $\mathbf{Monoids}\leftrightarrows\mathbf{Sets}$, $\mathbf{Mod}_R\leftrightarrows\mathbf{Sets}$ is certainly a good thing, but it would be slightly better to say that all these pairs come from a "general scheme of adjunction" $$ \text{generated object} \dashv \text{forgetful functor} $$ which can be (if I'm not wrong) studied for a general type of algebraic structure. Hence it would be better to write some sort of "reference card" about:

1. The diagonal functor $\Delta_\mathbf J\colon \mathbf C\to \mathbf C^\mathbf J$ sending $C\in\text{Ob}_\mathbf C$ into the constant diagram over $C$ admits both a left and right adjoint (direct and inverse limit).
2. Once you fixed a set $J$, here is an adjunction between $\mathbf{Sets}/J$ and $\mathbf{Sets}^J$ defined by functors $L\colon h\in \mathbf{Sets}/J\mapsto \big(h^\leftarrow(\{j\}\big)_{j\in J}$ and $M\colon \{H_j\}_{j\in J}\mapsto \big(\coprod_{j\in J} H_j\to J\big)\in \mathbf{Sets}/J$, which turns out to be an equivalence
3. There exists an adjunction between $\mathrm{PSh}(X)$ and $\mathbf{Top}/X$ for any topological space $X$ ($\text{bundle of germs}\dashv\text{(pre)sheaf of sections}$), which turns out to be an equivalence if we restrict...
4. Given a ring $R$ the functor $R[\;\;]\colon \mathbf{Groups}\to \mathbf{Rings}$ sending a group in its group ring admits a right adjoint, namely $U\colon R\mapsto R^\times$ (units in $R$).
5. The inclusion functor $\mathbf{Kelley}\to\mathbf{Top}$ admits a right adjoint, the kelleyfication of a topological space
6. (Following Gabriel&Zisman) The inclusion functor between (small) categories $\mathbf{cat}$ and (small) groupoids $\mathbf{Gpds}$, admits both a left adjoint ($\mathbf{C}\mapsto \mathbf{C}[\text{Mor}_\mathbf{C}^{-1}]$ in the notation used for the calculus of fractions) and a right adjoint ($\mathbf{C}\mapsto \mathbf{C}^\times$, sending a category in the groupoid obtained deleting every noninvertible arrow).
7. ...

Feel free to say this is a silly or boring question.

Answer 1

I'd like to mention a couple of non-standard examples.

1. The underlying set functor from $\mathbf{Top}$ to $\mathbf{Set}$ has a right adjoint (not just a left adjoint): the functor that endows every set with the indiscrete topology. The fact that the underlying set functor has adjoints on both sides is the reason behind the fact that for topological spaces, the underlying sets of all standard categorical constructions (products, coproducts, limits, colimits, etc) are the corresponding constructions of underlying sets; so not only is the underlying set of a product the (cartesian) product of the underlying sets, we also get that the underlying set of a coproduct is the coproduct (disjoint union) of the underlying sets.

2. Likewise, the forgetful functor from $\mathbf{Group}$ to $\mathbf{Monoid}$ has adjoints on both sides: the functor that maps a monoid to its group of units is the right adjoint of the forgetful functor, while the functor that sends the monoid to its universal enveloping group is the left adjoint.

3. Given any category $\mathbf{C}$ that has products and coproducts for all pairs, define $\mathbf{C}\times\mathbf{C}$ to be the category of all products $A\times B$, with $A,B\in\mathrm{Ob}(\mathbf{C})$, and arrows consisting of pairs $(f,g)\colon A\times B\to C\times D$, where $f\in\mathbf{C}(A,C)$ and $g\in\mathbf{C}(B,D)$. The diagonal functor $\Delta\colon\mathbf{C}\to\mathbf{C}\times\mathbf{C}$ sending $A$ to $A\times A$ and $f$ to $(f,f)$ has both a left and right adjoint: the right adjoint is the product functor, taking $(A,C)$ to $A\times C$; the left adjoint is the coproduct functor, taking $(A,C)$ to $A\amalg C$.

4. For a fairly naturally occurring category of algebras in which the underlying set functor has no adjoints, let $\mathbf{Div}$ be the category of divisible abelian groups. I claim that $\mathbf{Div}$ has no free objects.

   To see this, note the following:

   Proposition. Let $\mathbf{C}$ be a concrete category. If $\mathbf{C}$ has a free object in one generator, then monomorphisms in $\mathbf{C}$ are one-to-one functions.

   Proof. Let $f\colon A\to B$ be a monomorphism in $\mathbf{C}$. That means that for all objects $C$ and all morphisms $g,h\colon C\to A$, if $fg = fh$, then $g=h$ (i.e., $f$ is left-cancellable). Let $F(x)$ be the free object on one generator, $x$, and let $a,a'\in A$ be such that $f(a)=f(a')$. Let $g,h\colon F(x)\to A$ be the maps induced by the set theoretic maps $\mathfrak{g}\colon\{x\}\to A$ given by $\mathfrak{g}(x) = a$, and $\mathfrak{h}\colon\{x\}\to A$ given by $\mathfrak{h}(x)=a'$. Then $fg=fh$, hence $g=h$, hence $\mathfrak{g}\mathfrak{h}$, hence $a=a'$, proving that $f$ is one-to-one. QED

   To show that $\mathbf{Div}$ does not have free objects on one generator, consider the homomorphism $\mathbb{Q}\to\mathbb{Q}/\mathbb{Z}$. This is a monomoprhism in $\mathbf{Div}$: let $g,h\colon D\to\mathbb{Q}$ be such that $fg=fh$. Suppose $d\in D$ is such that $g(d)\neq h(d)$. Then $g(d)-h(d)=n\in\mathbb{N}$, $n\neq 0$; we may assume without loss of generality that $n\geq 1$. Let $x\in D$ be such that $(n+1)x=d$. Then $g(x)\neq h(x)$, and $$(n+1)(g(x)-h(x)) = g((n+1)x) - h((n+1)x) = g(d)-h(d) = n.$$ Therefore, $g(x)-h(x) = \frac{n}{n+1}\notin\mathbb{Z}$. But since $fg=fh$, then $f(g(x)-h(x)) = 0 + \mathbb{Z}$. This is impossible since $\frac{n}{n+1}\notin\mathbb{Z}$. The contradiction arises from the assumption that there exists $d\in D$ such that $g(d)\neq h(d)$, hence $g=h$. This proves that $f$ is a monomorphism.

   Since $f$ is a non-one-to-one monomorphism in $\mathbf{Div}$, it follows from the proposition that $\mathbf{Div}$ does not have free objects in one generator. Therefore, the underlying set functor $\mathbf{U}\colon\mathbf{Div}\to\mathbf{Set}$ does not have a left adjoint.

Answer 2

It's not the most important adjunction out there, but for me it was the example that drove home "no, really, they're everywhere".

Given a function $f:A\to B$ the inverse image functor $f^{-1}:\mathscr{P}(B)\to\mathscr{P}(A)$ (considering the powersets as the obvious preorder categories) has both a left and a right adjoint. Its left adjoint is the direct image function $f[-]$; its right adjoint Awodey refers to as the "dual image" that takes $a\in\mathscr{P}(A)$ to the largest subset of $b\in \mathscr{P}(B)$ with $f^{-1}[b]\subseteq a$.

It's easy to calculate, pretty, and can give a newcomer a taste of what adjoints do.

Answer 3

Here is an example from logic, and is an instance, in some sense, of the free-forgetful adjunction. Let $\textbf{Form}_\mathcal{L}(x_1, \ldots, x_n)$ be the poset category of first-order logical formulae over a fixed language $\mathcal{L}$ with $n$ free variables $x_1, \ldots, x_n$, with an arrow $p \to q$ if and only if $q$ may be derived from $p$. For simplicity we will assume that the names of bound variables are drawn from a second alphabet. Then, $\textbf{Form}_\mathcal{L}(x_1, \ldots, x_n)$ is a full subcategory of $\textbf{Form}_\mathcal{L}(t, x_1, \ldots, x_n)$, and the inclusion functor $U$ admits both a left and a right adjoint: $$(\exists t) \dashv U \dashv (\forall t)$$ Indeed, if $\varphi$ is an $n$-ary predicate and $\psi$ an $(n+1)$-ary predicate, $$\Updownarrow \frac{(\exists \alpha) \psi (\alpha, x_1, \ldots, x_n) \rightarrow \varphi (x_1, \ldots, x_n)}{\psi (t, x_1, \ldots, x_n) \rightarrow \varphi (x_1, \ldots, x_n)}$$ $$\Updownarrow \frac{\varphi (x_1, \ldots, x_n) \rightarrow \psi (t, x_1, \ldots, x_n)}{\varphi (x_1, \ldots, x_n) \rightarrow (\forall \alpha) \psi (\alpha, x_1, \ldots, x_n)}$$ Thus $\textbf{Form}_\mathcal{L}(x_1, \ldots, x_n)$ is a reflective and coreflective subcategory of $\textbf{Form}_\mathcal{L}(t, x_1, \ldots, x_n)$.

Answer 4

(Responding to a request...)

The forgetful functor from commutative $k$-algebras ($k$ itself a commutative ring) to sets has left adjoint the functor that forms the free commutative $k$-algebra on given sets. The case that the set is a singleton produces $k[x]$. (I like this example because it explains what an "indeterminate" is.)

Similarly, forgetful functors from groups to sets have left adjoints which form the corresponding free objects "on" a set.

For fields $k\subset K$ (or commutative rings...), for example the partly-forgetful ("restriction") functor from K-modules to k-modules has a left adjoint and a right adjoint, in general not the same, both called "extension of scalars". The left adjoint is M->$M\otimes_kK$, and the right adjoint is M->$Hom_k(K,M)$.

The analogous restriction functor for (e.g., finite) groups has both left and right adjoints, also, which do match for finite groups (maybe in characteristic zero). The adjoint to the forgetful functor is "induction", and the adjunctions relation is Frobenius Reciprocity.

In more general situations, say with p-adic groups, it can happen that one of the two Frobenius reciprocities is valid, and the other not! I saw this in Cartier's Corvallis article, and was further baffled, at the time, by the pains he went to to show that, nevertheless, the "other" induction functor was an adjoint of something (thereby proving its right exactness, was the point).

The ever-popular $Hom(A\otimes B,C)\approx Hom(A,Hom(B,C))$, say for abelian groups.

The functor "Lie" that takes an associative algebra A to the Lie algebra with bracket $[a,b]=ab-ba$ has left adjoint the map from a Lie algebra to its universal enveloping algebra.

The inclusion from sheaves to presheaves has left adjoint sheafification.

Global sections functor is right adjoint to the constant sheaf functor.

Stalk functor has right adjoint the skyscraper-sheaf functor.

Oop, almost forgot the tvs example: the forgetful functor from locally convex topological vector spaces to (algebraic) (complex) vector spaces has a left adjoint, which topologizes a vector space as the locally convex colimit of all the finite-dimensional subspaces (each of which has a unique tvs topology). Every linear map from this tvs is continuous. This example has a slight extra bit of amusement value because in the category of not-necessarily-locally-convex tvs's uncountable coproducts (e.g., of lines!) do not exist.

Answer 5

A very important adjunction for homotopy theory is the pair $$|\cdot|:\mathbf{sSet}\longleftrightarrow\mathbf{CGHaus}:S$$ between simplicial sets and the category of compactly generated Hausdorff spaces with the K-topology. The left adjoint is the geometric realization of a simplicial set (geometrically, you glue the basic simplices together along common faces), and the right adjoint is the simplicial chain functor.

The importance of this adjunction is in the fact that, when the categories are endowed with their standard model structures, it is a Quillen adjunction.

Answer 6

Consider a totally ordered set $(L,\leq)$ with the least upper bounded property, plus for each element $b$ of $L$, an operation $+b\colon L\to L$ so that $a_1\leq a_2$ implies $a_1+b\leq a_2+b$.

Then we can define an operation $-b$ sending each $c$ to $c-b=\sup\{d:d+b\leq c\}$, which exists because the totally ordered set has the least upper bound property (and when $+b$ has the additional property that $a_1+b=a_2+b$ implies $a_1=a_2$, then this gives the unique element $a$ so that $a+b=c$).

From a categorical point of view, the poset is a category with a unique morphism between any two elements, $+b$ is a functor from the category to itself, and $c-b$ is the reflection of $c$ along $+b$. Therefore, the operation $-b$, which is subtraction of $b$, is the left adjoint of the operation $+b$, which is addition of $b$. Furthermore, when $+b$ is the partial evaluation of a binary operation so that $b_1\leq b_2$ implies $a+b_1\leq a+b_2$, then the left adjoints assemble into a two-variable subtraction functor, covariant (order-preserving) in the first variable, and contravariant (order-reversing) in the second variable.

Answer 7

A very general kind of adjoint pairs of functors arising in algebra is the following:

Let $\mathcal{P}$ be an (algebraic) operad, $\mathcal{C}$ a cooperad, and let $\alpha$ be a twisting morphism from $\mathcal{C}$ to $\mathcal{P}$. Then $\alpha$ induces a bar-cobar adjunction $$B_\alpha:\text{dg $\mathcal{P}$-alg.}\longleftrightarrow\text{dg $\mathcal{C}$-coalg.}:\Omega_\alpha.$$ A standard example of this is the usual bar-cobar adjunction between (augmented) dg associative algebras and (conilpotent) cocommutative coalgebras.

Reference: B. Vallette, Homotopy Theory of Homotopy Algebras, page 5.

Answer 8

The inclusion functor $U \colon \mathbf{Set} \to \mathbf{Rel}$ has a right adjoint $\mathcal{P} \colon \mathbf{Rel} \to \mathbf{Set}$ which maps sets onto their power sets and relations $r \subset X \times Y$ onto $$ \mathcal{P}(r) \colon \mathcal{P}(X) \to \mathcal{P}(Y), \qquad S \mapsto \left\{ y \in Y : \exists \, s \in S \text{ such that } sry \right\}. $$ This is essentially the direct image functor but for relations instead of just functions. Its unit $\eta_X \colon X \to \mathcal{P}(U(X))$ is the singleton function $x \mapsto \{x\}$ and its counit $\varepsilon_Y \subset U(\mathcal{P}(Y)) \times Y$ is the reverse membership relation $S \; \varepsilon_Y \; y \Longleftrightarrow y \in S$ (coincidentally both of these relations are written with an epsilon symbol). Because products in $\mathbf{Rel}$ are disjoint unions and right adjoints are continuous this is a way to show that the Cartesian product of two power sets $\mathcal{P}(X) \times \mathcal{P}(Y)$ can be canonically identified with the power set of the disjoint union $\mathcal{P}(X \sqcup Y)$.