Could someone give some examples of a non-monadic adjunctions please? Possibly explaining why they are not monadic and how they contradict the monadicity theorem? Thanks!
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2$\begingroup$ According to nlab, the category of cancellative monoids provides an example. Explicitly, the adjunction consists of the usual forgetful functor to $\mathbf{Set}$, and its left-adjoint, which builds the usual free monoid. $\endgroup$– goblin GONEApr 21, 2016 at 15:03
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2 Answers
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A classical example : the forgetful functor from topological spaces to sets.
The left adjoint is the "discrete space" functor (sending a set $X$ to the discrete space with underlying space $X$), and the composition just gives the identity on Sets, so clearly Top is not the Eilenberg-Moore category of the monad.
You can see that the forgetful functor does not reflect isomorphisms (a homeomorphism is more than a bijective continuous function), so the monadicity theorem indeed cannot be applied.
answered Apr 21, 2016 at 14:48
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2$\begingroup$ Good one! This is a nice and intuitive example. And it emphasizes an important phenomenon, which is that our adjunctions tend to be monadic iff we're doing algebra. $\endgroup$ Apr 21, 2016 at 15:05
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4$\begingroup$ And magically, as soon as you pass to the spaces (compact Hausdorff) where $U$ does reflect isomorphisms, you have a monadic adjunction! Though not with the discrete space functor, of course. $\endgroup$ Apr 21, 2016 at 15:29
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This might not be the answer you are looking for, but it could give you a little insight on what monadic really means.
Given a monad $(T,\eta,\mu)$ on a category $\mathcal C$, you can form the category $\operatorname{Adj}(\mathcal C,T)$ of adjunctions above $T$:
- its objects are the ajdunction $F: \mathcal C \rightleftarrows \mathcal D :U$ ($F$ on the left) such that the induced monad $(UF,{\rm unit},U\,{\rm counit}\,F)$ equals $(T,\eta,\mu)$ ;
- a morphism from $F: \mathcal C \rightleftarrows \mathcal D :U$ to $F': \mathcal C \rightleftarrows \mathcal D' :U'$ is a functor $K \colon \mathcal D \to \mathcal D'$ such that $KF = F'$, $U'K = U$ and $K$ commutes with counits.
This category $\operatorname{Adj}(\mathcal C,T)$ has a terminal object: the Eilenberg-Moore category of $T$. The unique morphism from any other object is the so called comparison functor. Hence, by definition, a adjunction $F: \mathcal C \rightleftarrows \mathcal D :U$ is monadic if and only if it is terminal in $\operatorname{Adj}(\mathcal C,UF)$.
It should make you realize that, in a sense, most adjunctions are non-monadic. For example, the previous category $\operatorname{Adj}(\mathcal C,T)$ always have an initial object: the Kleisli category of $T$. And this category is terminal only if every algebra over $T$ is free! Which is quite uncommon. It produces a lot of examples of non-monadic adjunctions $$Set \rightleftarrows \text{Free algrebas over T}$$ for $T$ the monad of groups, monoids, modules over a non-field ring, etc.
Note that the initial and terminal object can be the same (as in the example of Captain Lama), but still leaving some "room" for non monadic adjunctions in between.
