A bestiary about adjunctions

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.

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.

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 $\textbf{Form}_\mathcal{L}(t, x_1, \ldots, x_n)$.