Though adjoint functors provide a universal description for many concrete mathematical constructions, these constructions usually revolve around finding a single "canonical" way to transform one type of object into another. For example, the only commonly mentioned pair of adjoint functors $(L \vdash R): \mathrm{Grp} \rightarrow \mathrm{Set}$ has $L$ as the free functor and $R$ as the forgetful functor, and the same seems to be true for any two categories where the notions of free and forgetful functor make sense. However, there is more than one "nice" pair of adjoint functors $(L \vdash R): \mathrm{Set} \rightarrow \mathrm{Cat}$; take $L$ to be the functor that forgets arrows and $R$ the indiscrete-category functor, or take $L$ to be the functor sending a category to its set of connected components (and acting similarly on arrows in $\mathrm{Cat}$) and $R$ the discrete-category functor.

Are there any more instances where there are two different pairs of adjoint functors (going in the same direction, so this would entail two non-isomorphic functors each of which have a right adjoint) with similarly "nice" descriptions between given inequivalent concrete categories? Especially interesting would be different adjoint pairs where there is already a free-forgetful adjunction.

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    $\begingroup$ Of course you can modify your example by replacing $\mathsf{Cat}$ by $\mathsf{Top}$. $\endgroup$ Commented Dec 7, 2014 at 21:25
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    $\begingroup$ Plenty more examples come from chains of adjoint functors $\dots \dashv F \dashv G \dashv H \dashv \dots$ here, here, and here. Many of these examples are concrete. $\endgroup$
    – tcamps
    Commented Dec 10, 2014 at 6:52

2 Answers 2


Sure. For example, there are (at least) two interesting forgetful functors from commutative rings to abelian groups: the underlying additive group, and the multiplicative group of units. Their left adjoints are, respectively, the symmetric algebra $A \mapsto S(A)$ and the group algebra $A \mapsto \mathbb{Z}[A]$.

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    $\begingroup$ Great example ! $\endgroup$ Commented Dec 7, 2014 at 20:50

Let $R \to S$ be a homomorphism of rings. Then $\mathsf{Mod}(R) \to \mathsf{Mod}(S)$, $M \mapsto M \otimes_R S$ (extension of scalars) is left adjoint to the forgetful functor (restriction of scalars). There is another functor $\mathsf{Mod}(R) \to \mathsf{Mod}(S)$, namely $\hom_R(S,-)$, which is right adjoint to the forgetful functor, but it is also left adjoint when $S$ is a f.g. projective (i.e. dualizable) $R$-module. In fact, then $\hom_R(S,-) \cong - \otimes_R S^*$ with right adjoint $\hom_S(S^*,-)$.

Another example is $S=R[\varepsilon]/(\varepsilon^2)$. The first left adjoint $\mathsf{Mod}(R) \to \mathsf{Mod}(R[\varepsilon]/(\varepsilon^2))$ is extension of scalars. There is another functor $\mathsf{Mod}(R) \to \mathsf{Mod}(R[\varepsilon]/(\varepsilon^2))$ which endows an $R$-module with the action $\varepsilon=0$. It is right adjoint to the functor which maps $M \mapsto M/\varepsilon M$. It is left adjoint to the functor which maps $M \mapsto \ker(\varepsilon : M \to M)$.


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