[Math] A bestiary about adjunctions

adjoint-functorsbig-listcategory-theory

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.

Best Answer

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.

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