Suppose $F,G$ are adjoint, and $\epsilon:F\circ G\rightarrow Id$ is the counit. Is it always true that$$
Id\_{FG}\epsilon=\epsilon Id\_{FG}
$$
as maps from $FGFG$ to $FG$?


It's true if you precompose with $F\eta G$, where $\eta:Id\rightarrow GF$ is the unit, then the result in both cases is $Id\_{FG}$, but since $\eta$ is not necessarily injective it doesn't imply what I want about the original maps.



Simple technical adjunction question
No, it is not always true. For instance, suppose $G$ is the forgetful functor from monoids to sets. Then $F$ is the free monoid functor, and the two maps in question are
$$((x\_{1,1}, \ldots, x\_{1,n\_1}), \ldots, (x\_{m,1}, \ldots, x\_{m,n\_m})) \mapsto (x\_{1,1} \cdots x\_{1,n\_1}, \ldots, x\_{m\_1} \cdots x\_{m,n\_m})$$
$$((x\_{1,1}, \ldots, x\_{1,n\_1}), \ldots, (x\_{m,1}, \ldots, x\_{m,n\_m})) \mapsto (x\_{1,1}, \ldots, x\_{1,n\_1}) \cdots (x\_{m\_1}, \ldots, x\_{m,n\_m})$$
More concretely, if we identify $F G 1$ with $\mathbb{N}$ and $F G F G 1$ with finite lists of natural numbers, then the two maps are (respectively)
$$(x\_1, \ldots, x\_m) \mapsto m$$
$$(x\_1, \ldots, x\_m) \mapsto x\_1 + \cdots + x\_m$$
which are obviously not the same.


It's worth pointing out that the equation holds if and only if $F$ and $G$ form an [idempotent adjunction](http://ncatlab.org/nlab/show/idempotent+adjunction). In particular, it is true when $G$ is fully faithful.


