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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?

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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 $Mnd(C)$.

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    $\begingroup$ So just need to figure out what a fibration object in Mnd is then. $\endgroup$
    – Max New
    Commented Dec 29, 2018 at 16:37
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    $\begingroup$ 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
    Commented Dec 30, 2018 at 15:00
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    $\begingroup$ 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
    Commented Dec 30, 2018 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
    Commented Jan 2, 2019 at 20:46
  • $\begingroup$ Oh wow! Yes, thanks for catching that! $\endgroup$
    – Tim Campion
    Commented Jan 4, 2019 at 15:16

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