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We know that, for any monad $T$, the Kleisli category $\mathcal{C}_T$ embeds into the Eilenberg-Moore category of $T$-algebras $\mathcal{C}^T$ as the full subcategory of free $T$-algebras. In the case of the monad for vector spaces, for example, this embedding is actually part of an equivalence of categories.

Are there nice categorical conditions on $T$, $\mathcal{C}_T$ or $\mathcal{C}^T$ that are sufficient to conclude that the full embedding $\mathcal{C}_T \hookrightarrow \mathcal{C}^T$ is essentially surjective, i.e that all $T$-algebras are essentially free?

José Siqueira
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    Doubtful. Note that the statement that all vector spaces are free is equivalent to the axiom of choice. – Qiaochu Yuan Jun 09 '17 at 15:46
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    I wouldn't be so sure. Note that the axiom of choice is equivalent to the fact that in both $\mathbf{Vec}$ and $\mathbf{Set}$ all epimorphisms split. – Contravariant Jun 09 '17 at 20:16
  • A sufficient condition is that the monad is idempotetn, right? – fosco Jun 09 '17 at 22:05
  • A necessary (but not sufficient) condition for monadicity which is not often mentioned is that the counit of the adjunction is a split epimorphism, which might be a useful thing to check. – Morgan Rogers Oct 29 '19 at 12:43

1 Answers1

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I think Beck's monadicity theorem provides an answer, it gives sufficient conditions for a functor $G: D \to C$ to be 'monadic' (i.e. has a left adjoint $F$ such that $D$ is the space of $GF$-algebras).

Applying this theorem to the forgetful functor $U : C_T \to C$ of the Kleisli, quite a few conditions are trivially satisfied, resulting in the following necessary and sufficient condition for the embedding $C_T \hookrightarrow C^T$ to be essentially surjective

$C_T$ has coequalizers of $U$-split parallel pairs and $U$ preserves those coequalizers

This condition is extremely technical, but if I understand the terms correctly a pair $f,g : A \to B$ has a split co-equaliser if there exists an arrow $h : B \to C$ such that the following commutes

$$ A \mathop{\rightrightarrows}^{f}_g B \mathop{\rightarrow}^h C $$ and there is a section $s$ of $h$ and a section $t$ of $f$ such that $g \circ t = s \circ h$.

Such a pair $f$, $g$ is called $U$-split if the above is true of $Uf$ and $Ug$.

Contravariant
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  • Thanks, I suppose that will have to do. At best, we can use other versions such as Crude Monadicity to try and make it simpler to check in some cases, but there is no running away from the Beck conditions --- we know $\mathcal{C}_T$ has coproducts when $\mathcal{C}$ does and that the forgetful functor from Kleisli has a left adjoint, but we can't say much automatically about coequalizers of any sort. – José Siqueira Jun 11 '17 at 10:44
  • I don't know if this is correct, but +1 because it sounds reasonable. – goblin GONE May 09 '19 at 11:32