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Stream: learning: questions

Topic: a monadicity theorem

Lucas Queiroz (he/him/his) (May 31 2022 at 02:57):

I would like to prove that if $U: \mathcal{D} \rightarrow \mathcal{C}$ is a functor with left and adjoint $F: \mathcal{C} \rightarrow \mathcal{D}$ and:

$\begin{enumerate}
\item $U$ reflects isomorphisms
\item $\mathcal{D}$ has coequalizers of reflexive pairs
\item $U$ preserves coequalizers of reflexive pairs
\end{enumerate}$

then the adjunction $F \vdash U$ is a monadic adjunction.

Under these assumptions, I can prove that the canonical comparison functor $K: \mathcal{D} \rightarrow \mathcal{C}^{UF}$ is essentially surjective. Given an algebra $\gamma: UFC \rightarrow C$, the parallel pair $\varepsilon_{FC}, F\gamma: FUFC \rightarrow FC$ is reflexive because $F\eta_C: FC \rightarrow FUFC$ is a section of both. It follows that there is a coequalizer $h: FC \rightarrow D$ for the pair $\varepsilon_{FC}, F\gamma$.

$Uh$ is coequalizer of $U\varepsilon_{FC}, UF\gamma$, but so is $\gamma$, hence there exists an isomorphism $\theta: C \rightarrow UD$ such that $Uh = \gamma; \theta$. To prove that $\theta$ is an algebra morphism $(C, \gamma) \xrightarrow{\theta} (UD, Uh)$ it suffices to check that $UF\gamma; \gamma; \theta = UF\gamma; UF\theta; U\varepsilon_{FC}$.

This shows that the comparison functor $K: \mathcal{D} \rightarrow \mathcal{C}^{UF}$ is essentially surjective. It remains to show that it is fully faithful.

Lucas Queiroz (he/him/his) (May 31 2022 at 02:59):

Is there a way to use latex here?

James Deikun (May 31 2022 at 03:12):

Surround with two $ signs instead of one. Or use a ```math block.

Mike Shulman (May 31 2022 at 05:04):

(I moved this to another topic.) I think the statement you're trying to prove is called the crude monadicity theorem. It's an immediate consquence of the precise monadicity theorem that gives necessary and sufficient conditions for a functor to be monadic, since its hypotheses are strictly stronger, so any reference that proves the precise monadicity theorem should show how to complete your proof as well. Is there a particular point you're stuck on?

Lucas Queiroz (he/him/his) (May 31 2022 at 10:17):

Mike Shulman said:

(I moved this to another topic.) I think the statement you're trying to prove is called the crude monadicity theorem. It's an immediate consquence of the precise monadicity theorem that gives necessary and sufficient conditions for a functor to be monadic, since its hypotheses are strictly stronger, so any reference that proves the precise monadicity theorem should show how to complete your proof as well. Is there a particular point you're stuck on?

By "precise monadicity theorem" do you mean the theorem "An adjunction $F \vdash U$ is monadic if and only if $U$ creates coequalizers of $U$-split pairs"?

Is the crude monadicity theorem a consequence of this theorem?

In the text have written above, I am half-way of proving the crude monadicity theorem, it only remains to prove that the comparison functor $K: \mathcal{D} \rightarrow \mathcal{C}^{UF}$ is fully faithfull, which unfortunately I do not know how to do right now.

Mike Shulman (May 31 2022 at 14:43):

I'm not sure if it's possible to deduce the crude monadicity theorem (CTT) directly from the precise one (PTT) stated in that way. But the U-split coequalizers that are used in the proof of PTT are, in fact, also reflexive; so the same proof of the "if" direction of PTT is a proof of CTT. Some references, like Toposes, triples, and theories, include reflexivity in the statement of PTT.

Lucas Queiroz (he/him/his) (May 31 2022 at 19:05):

Thank you for recommending the texts Toposes, triples, and theories. I am reading it now

Lucas Queiroz (he/him/his) (Jun 01 2022 at 02:57):

Lucas Queiroz said:

I would like to prove that if $U: \mathcal{D} \rightarrow \mathcal{C}$ is a functor with left and adjoint $F: \mathcal{C} \rightarrow \mathcal{D}$ and:

$\begin{enumerate}
\item $U$ reflects isomorphisms
\item $\mathcal{D}$ has coequalizers of reflexive pairs
\item $U$ preserves coequalizers of reflexive pairs
\end{enumerate}$

then the adjunction $F \vdash U$ is a monadic adjunction.

Under these assumptions, I can prove that the canonical comparison functor $K: \mathcal{D} \rightarrow \mathcal{C}^{UF}$ is essentially surjective. Given an algebra $\gamma: UFC \rightarrow C$, the parallel pair $\varepsilon_{FC}, F\gamma: FUFC \rightarrow FC$ is reflexive because $F\eta_C: FC \rightarrow FUFC$ is a section of both. It follows that there is a coequalizer $h: FC \rightarrow D$ for the pair $\varepsilon_{FC}, F\gamma$.

$Uh$ is coequalizer of $U\varepsilon_{FC}, UF\gamma$, but so is $\gamma$, hence there exists an isomorphism $\theta: C \rightarrow UD$ such that $Uh = \gamma; \theta$. To prove that $\theta$ is an algebra morphism $(C, \gamma) \xrightarrow{\theta} (UD, Uh)$ it suffices to check that $UF\gamma; \gamma; \theta = UF\gamma; UF\theta; U\varepsilon_{FC}$.

This shows that the comparison functor $K: \mathcal{D} \rightarrow \mathcal{C}^{UF}$ is essentially surjective. It remains to show that it is fully faithful.

@Mike Shulman I am still having some difficult in proving that $K: \mathcal{D} \rightarrow \mathcal{C}^{UF}$ is fully faithful. Can you give some hints please?

Mike Shulman (Jun 01 2022 at 14:57):

I'm sorry, I don't have the time to walk you through it more explicitly. Maybe someone else here will, or you can check some other references if TTT is too hard to follow.

Todd Trimble (Jun 01 2022 at 18:20):

See Mac Lane-Moerdijk, Sheaves in Geometry and Logic, for a direct proof.

Lucas Queiroz (he/him/his) (Jun 01 2022 at 23:18):

@Todd Trimble: I see, indeed Sheaves in Geometry and Logic proves the result I want on page 179. I will try to read this proof