2
$\begingroup$

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.

$\endgroup$

1 Answer 1

6
$\begingroup$

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. In particular, it is true when $G$ is fully faithful.

$\endgroup$
1
  • $\begingroup$ Just a small follow-up: It turned out my functors are not idempotent adjoint, so I had to rework my proof without using this, and now it's much more clear - so thanks again :-) $\endgroup$
    – Adam Gal
    Apr 1, 2014 at 22:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.