Suppose we have a category $C$ together with an endofunctor $L:C\to C$ and a natural transformation $\eta : Id_C\to L$. Suppose furthermore that the two natural transformations $L\eta, \eta L :L\to LL$ are isomorphisms.
Wikipedia claims that it follows that $L\eta = \eta L$, but I don't really see how it does, so my question is : is it the case, and if so, why ?
I tried to apply naturality to get the following commutative diagram :
$\require{AMScd} \begin{CD} A @>{\eta_A}>> LA\\ @V{\eta_A}VV @VV{L(\eta_A)}V\\ LA @>>{\eta_{LA}}> LLA \end{CD}$
So we do get a relation between $L\eta$ and $\eta L$, namely $L\eta \circ \eta = \eta L\circ \eta$, but I don't see how we could get much more. In the natural examples to come to mind, of course, this is satisfied; but I don't see how it would follow at this level of generality.
By using the same naturality diagram with $A=LB$, given that $\eta_{LB}$ is an isomorphism so we can simplify it, we get $L\eta L = \eta LL$; or if we take the same diagram with $A$ arbitrary, and apply $L$ to it, we can then simplify the $L(\eta_A)$'s so that $LL\eta = L\eta L$ ($=\eta LL$); but I really can't seem to get rid of there $L$ : maybe I'm missing something stupid.
A counterexample would (obviously) have $\eta_A$ be not epi. Also, if $L$ can be given a monad structure with unit $\eta$, then quite clearly (by the unit laws and the assumption that the two transformations are isomorphic) we do get $\eta L = L\eta$, as then $\mu : LL\to L$ is their common inverse. Perhaps that's why I can't find a counterexample : most of my examples of this situation are actually monads missing their $\mu$.
This made me think that perhaps I can show that under these hypotheses $L$ is a monad : consider the full subcategory $D$ of $C$ of objects that are isomorphic to some $LB$ (equivalently, under our hypotheses, those objects for which $\eta_A$ is an isomorphism), then perhaps we can prove that the inclusion $i:D\to C$ has a left adjoint such that the composition is $L$; and such that $\eta$ is the unit of the adjunction ?
Well of course our obvious functor is the corestriction of $L$ to $D$, which I will write $l$ to avoid (my own) confusion.
All of what follows is based on the following diagram :
$\require{AMScd} \begin{CD} A @>{f}>> B\\ @V{\eta_A}VV @VV{\eta_B}V\\ LA @>>{L(f)}> LB \end{CD}$
Then I have a map $\varphi_{A,B}:\hom(A,iB) \to \hom(LA,B)$ defined by $f\mapsto \eta_B^{-1}\circ L(f)$. This map is clearly natural in $(A,B)$; so let's see if it's a bijection. Surely $f= \varphi_{A,B}(f)\circ \eta_A$.
Conversely, $\varphi_{A,B}(g\circ \eta_A) = \eta_B^{-1}\circ L(g\circ \eta_A) = \eta_B^{-1}\circ L(g)\circ L(\eta_A)$. Ah ! but here I get stuck because to know that this is $g$ I would want to use the following diagram
$\require{AMScd} \begin{CD} LA @>{g}>> B\\ @V{L(\eta_A)}VV @VV{\eta_B}V\\ LLA @>>{L(g)}> LB \end{CD}$ but as far as I know, to prove that this is commutative I need to prove that $L\eta = \eta L$ so it's all circular.
Of course if $L$ is full I am done, because if $g:LA\to B$, then $\eta_B\circ g: LA\to LB$ is of the form $L(h)$ for some $h$ and then $\varphi_{A,B}(h) = g$. Conversely if my definition of $\varphi_{A,B}$ is surjective, then $L$ is full.
Therefore my question can be upgraded to :
Is such an $L$ always full ? Always a monad ? Does the equality always hold ?
The answer to "is it always full ?" is no, even for a monad (take $-\otimes \mathbb{F}_2$ on $\mathbf{Ab}$; then there is no nontrivial map $\mathbb{F}_2\to \mathbb{Z}$)
Clive pointed in the comments to a document which contains the answer: it turns out I just needed to go one step further.
Indeed I already know $LL\eta = L\eta L = \eta LL$ and we also have a commutative square
$\require{AMScd} \begin{CD} LA @>{L(\eta_A)}>> LLA\\ @V{\eta_{LA}}VV @VV{\eta_{LLA}}V\\ LLA @>>{LL(\eta_A)}> LLLA \end{CD}$
by naturality of $\eta$ and since $\eta_{LLA} = LL(\eta_A)$ and they're isomorphisms, it follows that $L(\eta_A) = \eta_{LA}$, i.e. $\eta L = L\eta$.
By the way, from this it also follows that $\varphi_{A,B}(g\circ \eta_A) = \eta_B^{-1}\circ L(g) \circ \eta_{LA} = g$ so that $\varphi_{A,B}$ is indeed bijective, with inverse$-\circ\eta_A$ and so we do have $L\dashv i$, in particular $iL$ (or $L$) is a monad on $C$, with unit $\eta$.