I'm familiar with a few examples of adjoint functor triples $F\dashv G\dashv H$:

  • equivalences of categories
  • left/right Kan extensions
  • induction/coinduction from a ring homomorphism
  • the forgetful functor from Top to Set

I'm curious if there are other simple examples I haven't heard about. Of the four listed, I actually find the last example most interesting, in both simplicity and that the two categories involved are actually very specific, rather than broad classes.

  • $\begingroup$ I just recalled an example I once knew - the forgetful functor from groupoids to categories. The left adjoint inverts all arrows, the right adjoint takes the core (subcategory of all isomorphisms). $\endgroup$
    – Christian
    Feb 22, 2021 at 18:36
  • $\begingroup$ If $G$ is a group, then the forgetful functor $G{-}\mathbf{Set} \to \mathbf{Set}$ has a left adjoint $X \mapsto G \times X$ where $G$ acts by left multiplication on the first coordinate; and it has a right adjoint $X \mapsto X^G$ where $G$ acts by $g \cdot \phi := \phi({-} \cdot g)$. However, this is very close to being a special case of the OP's third bullet point, using the ring homomorphism $\mathbb{Z} \to \mathbb{Z}[G]$, though that gives functors between abelian groups and abelian groups with actions of $G$. Still, close enough that I'll just leave it as a comment. $\endgroup$ Jul 10 at 19:55

2 Answers 2


There is a simple example between the category of graphs $\mathbf{Gph}$ and the category of sets $\mathbf{Set}$, see this post. To summarise it: we have the forgetful functor $U: \mathbf{Gph} \to \mathbf{Set}$. It has a left adjoint $F: \mathbf{Set} \to \mathbf{Gph}$, which assigns to a set $X$ the graph $F(X)$ with vertex set $X$ and no edges. Then there is also $C: \mathbf{Gph} \to \mathbf{Set}$ that sends a graph to the set of its connected components. We have $C \dashv F \dashv U$.

Interestingly, we can add a fourth functor to this list. Consider $G: \mathbf{Set} \to \mathbf{Gph}$ that sends a set $X$ to the graph $G(X)$ with vertex set $X$ and edges between any two vertices (and on functions it does nothing). Then $U$ is left adjoint to $G$.

A similar example would be to consider the category of preorders and order-preserving maps $\mathbf{PreOrd}$. Then the forgetful functor $U: \mathbf{PreOrd} \to \mathbf{Set}$ has both a left and right adjoint. Similar to the graphs example the left adjoint adds no order relations to our set, while the right adjoint adds all order relations.

It is worth noting that this last example fails if we would consider the category of partial orders instead. In fact, the forgetful functor from partial orders to the category of sets has no right adjoint, see this post.

  • $\begingroup$ The graphs example is a great one! I actually don't know anything about the category of simple graphs - is it involved in any other noteworthy adjunctions? $\endgroup$
    – Christian
    Feb 22, 2021 at 14:08
  • $\begingroup$ @Christian I would not know of any from the top of my head. But that does not mean there aren't any! $\endgroup$ Feb 22, 2021 at 16:53
  • $\begingroup$ I think you need to allow graphs to have self-loops (and make $G$ add these self-loops) if you want $G$ to be the right adjoint to $U$. $\endgroup$ Feb 23, 2021 at 2:56

There is a series of adjunctions between several functors of interest between $\mathbf{Ab}$ and $\mathbf{Ab}^{\rightarrow}$, where the second is the category where an object is a homomorphism $X \to Y$ of abelian groups, and a morphism is a commutative square. From left to right, these functors are:

  • $\operatorname{coker} : \mathbf{Ab}^{\rightarrow} \to \mathbf{Ab}$.
  • $\mathbf{Ab} \to \mathbf{Ab}^{\rightarrow}, X \mapsto (0 \to X)$.
  • $\operatorname{codom} : \mathbf{Ab}^{\rightarrow} \to \mathbf{Ab}, (X \to Y) \mapsto Y$.
  • $\mathbf{Ab} \to \mathbf{Ab}^{\rightarrow}, X \mapsto (X \overset{\operatorname{id}_X}{\rightarrow} X)$.
  • $\operatorname{dom} : \mathbf{Ab}^{\rightarrow} \to \mathbf{Ab}, (X \to Y) \mapsto X$.
  • $\mathbf{Ab} \to \mathbf{Ab}^{\rightarrow}, X \mapsto (X \to 0)$.
  • $\operatorname{ker} : \mathbf{Ab}^{\rightarrow} \to \mathbf{Ab}$.

Therefore, any element of this list other than the endpoints will be an answer to the original question.

(In fact, this will work for any abelian category in place of $\mathbf{Ab}$. And then, the symmetry of this list is related to the fact that the opposite category of an abelian category is also an abelian category, and the functor $X \mapsto \operatorname{id}_X$ is self-dual.)


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