The Wikipedia page for Monad says just that for a monad $(T,\eta,\mu)$ we can define the category of all adjunctions that define the monad:

Let $\textbf{Adj}(C,T)$ be the category whose objects are the adjunctions $(F,G,e,\varepsilon)$ such that $(GF,e,G\varepsilon F)=(T,\eta,\mu)$ and whose arrows are the morphisms of adjunctions which are the identity on $C$. Then this category has

  • an initial object $(F_T,G_T,\eta,\mu_T) : C\to C_T$, where $C_T$ is the Kleisli category,
  • a terminal object $(F^T,G^T,\eta,\mu^T) : C\to C^T$, where $C^T$ is the Eilenberg-Moore category.

What are the morphisms of adjunctions in $\textbf{Adj}(C,T)$ and what is meant by ... which are the identity on $C$?


1 Answer 1


The definition of morphism of adjunctions may be found in MacLane's book. Let $F:\mathcal C\rightarrow\mathcal D$, $G:\mathcal D\rightarrow\mathcal C$, $F':\mathcal C'\rightarrow\mathcal D'$, $G':\mathcal D'\rightarrow\mathcal C'$, and let $(F,G,\eta,\varepsilon)$, $(F',G',\eta',\varepsilon')$ be adjunctions. A map of adjunctions from the first to the second is a pair of functors $L:\mathcal C\rightarrow\mathcal C'$, $K:\mathcal D\rightarrow\mathcal D'$ such that $KF=F'L$, $LG=G'K$ and $L\eta=\eta'L$ (or, equivalently, $\varepsilon K=K\varepsilon'$).

In the case of the category of adjunctions of a monad, the category $\mathcal C$ on which the endofunctor of the monad acts is fixed, so $\mathcal C'=\mathcal C$ and one considers the above definition only in case $L=Id_{\mathcal C}$.

  • $\begingroup$ I'm a bit disturbed by the $=$ sign between functors: surely this means naturally isomorphic? $\endgroup$
    – cody
    Commented Apr 3, 2015 at 17:20
  • 1
    $\begingroup$ @cody: MacLane's definition (Section IV.7, p.97) uses equality. That seems to be the standard definition (see for instance here). I don't know what the consequences of a relaxed definition would be. $\endgroup$ Commented Apr 3, 2015 at 18:03

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.