6
$\begingroup$

Let $\mathrm{F}: \mathcal{C} \rightleftarrows \mathcal{D} : \mathrm{G} $ be an adjunction with associated monad $\mathrm{T} = \mathrm{G} \mathrm{F} .$

If $\mathcal{D} $ admits coequalizers of $\mathrm{G} $-split pairs, then the comparison functor $ \bar{\mathrm{G}}: \mathcal{D} \to \mathrm{Alg}_{ \mathrm{T}} (\mathcal{C}) $ admits a left adjoint $ \bar{\mathrm{F}}.$

So we obtain an adjunction $ \bar{\mathrm{F}}: \bar{ \mathcal{C} }:= \mathrm{Alg}_{ \mathrm{T}} (\mathcal{C}) \rightleftarrows \mathcal{D} : \bar{\mathrm{G}} $ with associated monad $\bar{\mathrm{T}} = \bar{\mathrm{G}} \bar{\mathrm{F}}.$

If $\mathcal{D} $ admits coequalizers of $\bar{\mathrm{G}} $-split pairs, then the comparison functor $ \mathcal{D} \to \mathrm{Alg}_{ \bar{\mathrm{T}}} (\bar{ \mathcal{C} }) $ admits a left adjoint.

So if we assume that $\mathcal{D} $ admits coequalizers (in fact reflexive coequalizers are enough), we can iterate this process starting with an adjunction $\mathrm{F_1}: \mathcal{C}_1 \rightleftarrows \mathcal{D} : \mathrm{G}_1 $ and get a sequence $ (\mathrm{F_\mathrm{i}}: \mathcal{C}_\mathrm{i}\rightleftarrows \mathcal{D} : \mathrm{G}_\mathrm{i})_{ \mathrm{i} \geq 1 } $ of adjunctions, where $\mathrm{G}_\mathrm{i}: \mathcal{D} \to \mathcal{C}_\mathrm{i}:= \mathrm{Alg}_{ \mathrm{T}_{\mathrm{i}-1} } ( \mathcal{C}_{\mathrm{i}-1} ) $ is the comparison functor for the adjunction $ \mathrm{F_{\mathrm{i}-1}}: \mathcal{C}_{\mathrm{i}-1}\rightleftarrows \mathcal{D} : \mathrm{G}_{\mathrm{i}-1} $ with associated monad $\mathrm{T}_{\mathrm{i}-1} $ if $\mathrm{i} > 1.$

Write $\mathcal{C}_{\infty}$ for the limit of the diagram $ ... \to \mathcal{C}_3 \to \mathcal{C}_2 \to \mathcal{C}_1$ so that we obtain a functor $\mathcal{D} \to \mathcal{C}_{\infty}$.

The functor $\mathcal{C}_{\infty} \to \mathcal{C}_1$ is right adjoint and conservative.

If $\mathcal{D}$ and all the categories $ \mathcal{C}_{\mathrm{i}} $ are presentable, all the right adjoint functors $ \mathrm{G}_{\mathrm{i}} $ preserve small limits and are accessible so that $\mathcal{D} \to \mathcal{C}_{\infty}$ preserves small limits and is accessible and so admits a left adjoint by the adjoint functor theorem.

Does the functor $\mathcal{D} \to \mathcal{C}_{\infty}$ admit a left adjoint without the presentability assumption on $\mathcal{D}?$

Denote $\mathrm{T}_\infty$ the monad associated to the adjunction $\mathcal{C}_{\infty} \rightleftarrows \mathcal{D}.$

It is tempting to believe that the monadic forgetful functor $\mathrm{Alg}_{\mathrm{T_\infty}} ( \mathcal{C}_\infty) \to \mathcal{C}_\infty $ is an equivalence. This is equivalent to the condition that $\mathrm{T_\infty}$ is the identity monad or that the left adjoint of the functor $\mathcal{D} \to \mathcal{C}_\infty $ is fully faithul. Is this true?

What can one say more about $\mathcal{C}_{\infty}$ and the functor $\mathcal{D} \to \mathcal{C}_{\infty}$?

$\endgroup$
7
$\begingroup$

These questions were studied (in the dual case of comonads, née cotriples) by Applegate and Tierney in their 1970 paper Iterated cotriples.

The answer to your first question is that yes, this functor always has an adjoint as long as $\mathcal{D}$ is cocomplete. The adjoint can be constructed using a sequential colimit in $\mathcal{D}$ of the individual adjoints $\mathcal{C}_i\to \mathcal{D}$.

In general I expect that the adjoint $\mathcal{C}_\omega \to \mathcal{D}$ is not fully faithful, although A&T don't give a counterexample. But they do show that the construction can be continued to all transfinite stages, and that if $\mathcal{D}$ is well-copowered then there is a sensible "$\Omega$-indexed limit" of the tower (where $\Omega$ is the well-ordered proper class of ordinals) for which the analogous adjoint exists and is fully faithful.

Edit: This decomposition was studied further by Adamek-Herrlich-Tholen in Monadic decompositions and by MacDonald-Stone in The tower and regular decomposition. I believe one or both of them may contain counterexamples to convergence at $\omega$.

$\endgroup$
1
$\begingroup$

@MikeShulman’s answer gives the main references, but here is a quick counterexample to your specific question about whether the monad $T_\infty$ must be the identity. Say a widget is a set $X$ equipped with $\mathbb{N}$-many partial operations $f_i : X \rightharpoonup X$, such that:

  • each $f_i$ is an endomorphism of a subset $X_i \subseteq X$;
  • $X_0$ is the whole of $X$;
  • $X_{i+1}$ is precisely the set of fixpoints of $f_i$.

Say a widget is special if it satisfies the following axiom: if there is any element which is fixed by all $f_i$, then all elements of $X$ are equal.

Now widgets are finitary essentially-algebraic structures, and specialness is an infinitary essentially-algebraic condition; so there’s an evident notion of map between widgets — i.e. a map preserving the operations whenever they’re defined — and the categories of these structures and maps are very well-behaved. Write $\newcommand{\W}{\mathbf{W}}\W$ for the category of widgets; $\W^\mathrm{sp}$ for the category of special widgets; and $\W_n$ for the category of objects like widgets, but with only the first $(n-1)$-many operations.

Resolving the adjunction of the forgetful functor $\W^\mathrm{sp} \to \newcommand{\Set}{\mathbf{Set}}\Set$, I claim we get exactly the tower of forgetful functors $\cdots \to \W_2 \to \W_1 \to \Set$; and the limit of this tower is just $\W$. But now the monad of the forgetful functor $\W^\mathrm{sp} \to \W$ is not the identity — in fact, this adjunction is monadic.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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