Say we have a Grothendieck fibration $p : E \to B$ and a monad $T$ on $B$ and a lift $T'$ of $T$ to $E$, i.e. a monad on $E$ such that $pT' = Tp$ and $p$ preserves $\eta, \mu$.

Then because the Eilenberg–Moore construction is functorial, we have a morphism $EM(p)$ from $T'\text{-}Alg$ to $T\text{-}Alg$. Is $EM(p)$ generally a fibration? If not, under what conditions is $EM(p)$ a fibration?


Let $C$ be a 2-category and $Mnd(C)$ the 2-category of monads in $C$. As explained by Street in the formal theory of monads, the Eilenberg-Moore construction is right 2-adjoint to the inclusion 2-functor $C \to Mnd(C)$ sending an object to the identity monad on it. Therefore it preserves 2-limits.

A fibration in a 2-category may be defined in terms of certain 2-limits (namely slice categories) and adjoints, and so is preserved by any 2-functor preserving 2-limits, and in particular by the Eilenberg-Moore construction.

Thus it's sufficient for $p$ to be a fibration object in the 2-category $Mnd(C)$.

  • 1
    $\begingroup$ So just need to figure out what a fibration object in Mnd is then. $\endgroup$ – Max New Dec 29 '18 at 16:37
  • 1
    $\begingroup$ Fibrations can be detected by homming in, so one answer will probably be a tautological one: $p: E \to B$ is a fibration if $Mnd(C)(X,E) \to Mnd(C)(X,B)$ is a fibration for all monads $X$, which will probably reduce in the case $C = Cat$ to the induced map of Eilenberg-Moore objects being a fibration. $\endgroup$ – Tim Campion Dec 30 '18 at 15:00
  • 2
    $\begingroup$ I think it would probably be easier to understand fibrations in Mnd(C) using the characterization of fibrations in terms of limits. Since the forgetful functor $\mathit{Mnd}(C)\to C$ also preserves limits (when C has Kleisli objects it has a left adjoint), being a fibration in Mnd(C) just means being a fibration in C together with the extra structure that the universal lift-assigning functor and transformation lift to Mnd(C). Then just work out what that means explicitly. $\endgroup$ – Mike Shulman Dec 30 '18 at 15:42
  • $\begingroup$ When you say the forgetful functor $Mnd(C) \to C$ do you mean the inclusion $C \to Mnd(C)$ that picks the identity monad instead? That's what it looks like it says in Street. $\endgroup$ – Max New Jan 2 at 20:46
  • $\begingroup$ Oh wow! Yes, thanks for catching that! $\endgroup$ – Tim Campion Jan 4 at 15:16

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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