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Suppose we have an adjoint triple $F \dashv G \dashv H$ with the following (co)units: $$\eta : I \to GF, \ \epsilon \colon FG \to I, \ \bar{\eta} : I \to HG, \ \bar{\epsilon} \colon GH \to I.$$ Consider the diagram

$$\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} FGH & \ra{F \bar{\epsilon}} & F \\ \da{\epsilon H} & & \da{\bar{\eta} F} \\ H & \ras{H \eta} & HGF \end{array}$$

Question: Is this diagram commutative?

If necessary, we can assume that $G$ is fully faithfull, which implies that $\bar{\eta}$ and $\epsilon$ are isomorphisms.

I belive it might be possible to construct cube-diagram from triagle identities and naturality conditions on (co)units, such that uper base of that cube is our considered diagram, but I haven't been able to do that.

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    $\begingroup$ When $G$ is fully faithful, it is $\epsilon$ and $\bar{\eta}$ that are natural isomorphisms. $\endgroup$
    – Zhen Lin
    Commented Jan 10, 2014 at 15:46

2 Answers 2

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First, a counterexample when $G$ is not fully faithful: take $G$ to be the diagonal functor $\Delta : \mathbf{Set} \to \mathbf{Set}$; then its left adjoint is disjoint union and its right adjoint is cartesian product, and the two canonical maps $(X \times Y) \amalg (X \times Y) \to (X \amalg Y) \times (X \amalg Y)$ are different in general.

On the other hand, it is true when $G$ is fully faithful. In fact, it is enough to assume that $G \bar{\eta}$ is a natural isomorphism. Then, $(G \bar{\eta})^{-1} = \bar{\epsilon} G$, so $G \bar{\eta} \bullet \bar{\epsilon} G = \mathrm{id}_{G H G}$, and: \begin{align} H \eta \bullet \epsilon H & = \epsilon H G F \bullet F G H \eta \\ & = \epsilon H G F \bullet (F G \bar{\eta} F \bullet F \bar{\epsilon} G F) \bullet F G H \eta \\ & = (\bar{\eta} F \bullet \epsilon F) \bullet (F \eta \bullet F \bar{\epsilon}) \\ & = \bar{\eta} F \bullet F \bar{\epsilon} \end{align}

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  • $\begingroup$ The first and third = come from the interchange law, right? $\endgroup$ Commented Jan 10, 2014 at 16:57
  • $\begingroup$ Yes, or more simply, by naturality. $\endgroup$
    – Zhen Lin
    Commented Jan 10, 2014 at 17:35
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If $R \to S$ is a ring homomorphism, we have an adjoint triple $$- \otimes_R S \dashv U \dashv \hom_R(U(S),-),$$ where $U : \mathsf{Mod}(S) \to \mathsf{Mod}(R)$ is the forgetful functor. Your diagram evaluated at some $M \in \mathsf{Mod}(R)$ looks as follows (I will omit $U$ from the notation): $$\begin{array}{cc} \hom_R(S,M) \otimes_R S & \rightarrow & M \otimes_R S \\ \downarrow && \downarrow \\ \hom_R(S,M) & \rightarrow & \hom_R(S,M \otimes_R S)\end{array}$$ It maps $f \otimes s$ to $f(1) \otimes s \in M \otimes_R S$, which is mapped to $t \mapsto f(1) \otimes s t$ in $\hom_R(S,M \otimes_R S)$. In the other direction, it maps $f \otimes s$ to $t \mapsto f(st)$ in $\hom_R(S,M)$, which is then mapped to $t \mapsto f(st) \otimes 1$ in $\hom_R(S,M \otimes_R S)$. From this we see that the diagram isn't commutative in general.

Moreover, it is commutative for all $M$ iff $s \otimes 1 = 1 \otimes s$ for all $s \in S$, which is equivalent to $R \to S$ being an epimorphism, which is equivalent to the fully faithfulness of $U$. After seeing Zhen Lin's answer, this is no surprise.

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