category theory

## Applications

#### Higher algebra

higher algebra

universal algebra

## Theorems

#### 2-Category theory

2-category theory

# Contents

## Idea

A functor is monadic if it is equivalent to the forgetful functor from a category of algebras over a monad. In this case it is part of a monadic adjunction.

## Definition

Given a pair of adjoint functors $F: C \to D :U$, $F \dashv U$, with unit $\eta: Id_C \to U \circ F$ and counit $\epsilon: F \circ U \to Id_D$, one constructs a monad $\mathbf{T}=(T,\mu,\eta)$ setting $T = U \circ F: C \to C$, $\mu = U \epsilon F: T T = U F U F \to U F = T$.

Consider the Eilenberg–Moore category $C^{\mathbf{T}}$ of $T$-algebras ($T$-modules) in $C$. Clearly $U (\epsilon_M): T U M = U F U M \to U M$ is a $T$-action. In fact there is a canonical comparison functor $K^{\mathbf{T}}: D \to C^{\mathbf{T}}$ given on objects by $K(M)=(U M, U (\epsilon_M))$. We then say that we have a monadic adjunction.

A functor $U: D \to C$ is monadic (resp. strictly monadic) if it has a left adjoint $F: C\to D$ and the comparison functor $K^{\mathbf{T}}: D \to C^{\mathbf{T}}$ is an equivalence of categories (resp. an isomorphism of categories). In other words, up to equivalence, monadic functors are precisely the forgetful functors defined on Eilenberg–Moore categories for monads, and strictly monadic functors are the same as these forgetful functors up to isomorphism. A category $D$ is monadic over a category $C$ if there is a functor $U: D \to C$ which is monadic.

## Properties

Various versions of Beck’s monadicity theorem (old-fashioned name of some schools: tripleability theorem) give sufficient, and sometimes necessary, conditions for a given functor to be monadic. There are also dual, comonadic versions.

A monadic functor is strictly monadic if and only if it is also an amnestic isofibration: clearly, a strictly monadic functor is an amnestic isofibration; and if a monadic functor $U$ is amnestic, then the comparison functor $K$ is also amnestic, and if $U$ is a monadic isofibration, so is $K$; therefore in this case $K$ must be an isomorphism of categories.

Revised on July 23, 2014 05:20:52 by Urs Schreiber (89.204.154.176)