Do adjoint functors really define monads?

Question

It is often claimed as "obvious" that a pair of adjoint functors: $L\colon{\cal V}\to {\cal M}$ and $R\colon{\cal M}\to {\cal V}$ defines a cotriple $(\bot, \epsilon, \delta)$ and a monad. What is wrong with the following counterexample?

Let ${\cal V}$ be the category of vector spaces over $\mathbb R$ and ${\cal M}$ be the category of modules over a Lie algebra $\mathfrak g.$ Then (I believe) the forgetful functor $R:{\cal M}\to {\cal V}$ is right adjoint to $L\colon{\cal V}\to {\cal M},$ $L(V)=V\otimes \mathfrak g.$

The counit $\epsilon\colon M\otimes \mathfrak g\to M$ is given by the action of $\mathfrak g$ on $M.$ In Weibel's "Intro to Homological Algebra", the construction of a monad is based on the following identity: $\epsilon\circ (LR\epsilon)=\epsilon\circ(\epsilon LR)$. That means that $(m\cdot x)\cdot y=m\cdot (x\cdot y)$ for $x,y\in \mathfrak g.$ But for Lie algebra modules we have $(m\cdot x)\cdot y=m\cdot (x\cdot y)+(m\cdot y)\cdot x$!

Answer 1

I think I found the error: My $L$ and $R$ are not adjoint! For an associative algebra $A,$ over $\mathbb R$, the forgetful functor $R$ is right adjoint to $L: {\mathcal V}\to {\mathcal M}$ (= category of right $A$-modules) by sending for example $z\in M=Hom_{\mathbb R}({\mathbb R},M)$ to $r_z\in Hom_{A}(A,M),$ $r_z(a)=z\cdot a.$ This is an $A$-module homomorphisms because $r_z(ab)=r_z(a)b$. But that fails when $A$ is not associative!

Thank you all for your comments, especially @Omar who pointed in that direction!

Answer 2

I've edited the answer to address the real problems of the question and to shorten it a little bit.

The data you have indicated cannot give an adjunction, the problem being in the supposed counit. The details follows.

The point is the the morphisms $\epsilon \colon M \otimes \mathfrak g \to M$ in order to be the (components of the) counit of the adjunction should be morphism in the category $\mathcal M$, i.e. they should be morphisms of lie algebras. And that's not possible here's why.

Ofcourse since the action of $\mathfrak g$ over a module $M$ is a bilinear map we can identify the action with a linear map $\epsilon_M \colon M \otimes \mathfrak g \to M$ which is the mapping such that $$\epsilon_M(m \otimes x) =m \cdot x$$ for $m \in M$ and $x \in \mathfrak g$, what we said until now tells that $\epsilon$ should satisfy the equality $$\epsilon_M( (m \otimes x) \cdot y) = \epsilon(m \otimes x) \cdot y$$ where the action $\cdot$ on the left is the action in the $\mathfrak g$-module $M \otimes \mathfrak g$ while in the action on the right is that in the module $M$.

Now the last equation could be rewritten as $$m \cdot [x,y] = (m \cdot x)\cdot y$$ that as you said in the question doesn't hold. So the $\epsilon$ is not a family of morphisms in $\mathcal M$ and so cannot be the counit of an a adjunction.

Just for completeness and to convince you that the property of adjoint functors that you stated, namely that the equality $\epsilon \circ LR(\epsilon) = \epsilon \circ \epsilon_{LR}$, holds I've written a proof below.

A little notation: $F \colon \mathcal X \to \mathcal A$ and $R \colon \mathcal A \to \mathcal X$ are the adjoint functors, $\varphi \colon \mathcal A(L(-),-) \cong \mathcal X(-,R(-))$ is the adjunction and $\epsilon \colon LR \Rightarrow 1_\mathcal{A}$ is the counit.

Now by the properties of adjunction we get that for every object $A \in \mathcal A$ $$\epsilon_A \circ \epsilon_{LR(A)}=\mathcal A(L(1_{R(M)}),\epsilon_A)\circ \varphi^{-1}(1_{RLR(A)})$$ $$\epsilon_A \circ \epsilon_{LR(A)} = \varphi^{-1} \circ \mathcal V(1_{R(A)},R(\epsilon_A))(1_{RLR(A)})$$ $$\epsilon_A \circ \epsilon_{LR(A)} = \varphi^{-1}(R(\epsilon_A))$$ and that $$\epsilon_A \circ LR(\epsilon_A) = \mathcal M(LR(\epsilon_A),1_{LR(A)}) \circ \varphi^{-1}(1_{R(A)})$$ $$\epsilon_M \circ LR(\epsilon_A) = \varphi^{-1}\circ \mathcal V(R(\epsilon_A),R(1_{LR(A)}))(1_{R(A)})$$ $$\epsilon_A \circ LR(\epsilon_A) = \varphi^{-1}(R(\epsilon_A))$$ this proves the so wished equality.

Answer 3

I would leave this as a comment, but see Mac Lane's "Categories for the Working Mathematician", second edition, p.138. He does not leave it at "it's obvious", but explains very clearly.

EDIT: Also, it took me a bit to notice this, but draw out the identity $\epsilon\circ LR\epsilon = \epsilon\circ \epsilon LR$ as a commutative diagram and observe that it's a naturality square for $\epsilon$. If your "counterexample" holds, $\epsilon$ isn't even a natural transformation, and thus not the counit of an adjunction.