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Is it possible to have two non-equivalent categories $C, D$ and functors $F : C \to D, G : D \to C$ with $$ F \dashv G \dashv F \dashv G$$ ?

Here $F \dashv G$ means that $F$ is left adjoint to $G$.

I know only two examples of functors having both left and right adjoints: $Top \to Set$ and $Mod_S \to Mod_R$ as here. But the adjoints are not the same functor in these cases. I added "non-equivalent" to avoid $C=D, F=G=id$.

Alphonse
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1 Answers1

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By coincidence, I recently stumbled upon the following example in a paper of Bourn : take any category $\mathcal{C}$, and consider the category $Pt(\mathcal{C})$ of pointed objects in $\mathcal{C}$. It is defined as the category whose

  • objects are quadruples $(A,B,p,s)$ such that $A,B$ are objects in $\mathcal{C}$ and $p:A\to B$ and $s:B\to A$ are morphisms such that $p\circ s=id_B$ in $\mathcal{C}$;
  • morphisms are commutative square between the split epimorphisms, i.e. pair of morphisms $\alpha : A\to A'$ and $\beta:B\to B'$ such that $\beta\circ p=p'\circ \alpha$ and $\alpha\circ s=s'\circ \beta$ in $\mathcal{C}$.

Then you have a functor $\operatorname{cod}:Pt(\mathcal{C})\to \mathcal{C}$ that sends $(A,B,p,s)$ to $B$, and a functor $\operatorname{dis}:\mathcal{C}\to Pt(\mathcal{C})$ that sends an object $C$ to $(C,C,id_C,id_C)$. Then we have $\operatorname{dis}\dashv \operatorname{cod}$, because in a commutative square $$\require{AMScd} \begin{CD}C@>>> A \\ @A{id_C}AA @AA{s}A \\ C@>>{f}> B\end{CD}$$ the top map can only be $s\circ f$, so to give a morphism $\operatorname{dis}(C)\to (A,B,p,s)$ is the same thing as giving a morphism $C\to \operatorname{cod}(A,B,p,s)$; and in the same way, in a commutative square $$\require{AMScd} \begin{CD}A @>>> C \\ @V{p}VV @VV{id_C}V \\ B @>>{g}> C\end{CD}$$ the top map can only be $g\circ p$, so we also have $\operatorname{cod}\dashv \operatorname{dis}$.

In fact, this is a special case of a more general construction. Given a category $\mathcal{X}$ with a zero object, you can consider two functors :

  • the functor $\Delta:\mathcal{C}\to [\mathcal{X},\mathcal{C}]$ that sends an object $C$ to the constant functor $\Delta_C$;

  • the functor $E_0:[\mathcal{X},\mathcal{C}]\to \mathcal{C}$ that evaluates a given functor $\mathcal{X}\to\mathcal{C}$ at the zero object of $\mathcal{X}$.

Since the zero object is initial, this functor $E_0$ amounts to taking the limit of a functor $\mathcal{X}\to\mathcal{C}$, so it is right adjoint to $\Delta$; since it is also terminal, it gives the colimit (as explained in the answer to this question), so it is also left adjoint to $\Delta$.

The example described above is the specific case where $\mathcal{X}$ is the category with two objects $X_1,X_0$ and three non-identity maps $p:X_1\to X_0$, $s:X_0\to X_1$ and $s\circ p:X_1\to X_1$, where $p\circ s=id_{X_0}$, so it is a sense the smallest non-trivial example of this form.

Arnaud D.
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