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Let $T = (t, \mu, \eta)$ be a monad on an object $A$ of a 2-category $\mathcal K$. In The formal theory of monads, Street proves (Theorem 3) that if $l \dashv r$ is the canonical adjunction associated to an Eilenberg–Moore object, then $l \dashv r$ is terminal amongst adjunctions inducing $T$. Is there a reference or simple proof of the converse, i.e. that if $l \dashv r$ is terminal amongst adjunctions inducing $T$, then $r$ exhibits an Eilenberg–Moore object for the induced monad? Alternatively, is there a counterexample?


Presumably the following universal property is the key, but I don't see quite how the proof would go. An Eilenberg–Moore object for $T$ has the universal property that $\mathcal K[X, A^T] \cong \mathcal K[X, A]^{\mathcal K[X, T]}$ 2-natural in $X$ (Theorem 8). Consequently, an adjunction $l \dashv r$ is the canonical adjunction associated to an Eilenberg–Moore object (cf. Theorem 2) if and only if $\mathcal K[X, l] \dashv \mathcal K[X, r]$ is monadic in $\mathbf{Cat}$ in the usual sense (Corollary 8.1). Supposing that $l \dashv r$ is terminal, we wish to show that $\mathcal K[X, l] \dashv \mathcal K[X, r]$ is terminal. However, applying $\mathcal K[X, {-}]$ does not seem sufficient, since there may be adjunctions inducing $\mathcal K[X, T]$ that are not of the form $\mathcal K[X, l'] \dashv \mathcal K[X, r']$, in which case we cannot apply terminality of $l \dashv r$.

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I don't know the answer to the original question, but I don't think I'd be alone in arguing that the condition "there is an adjunction terminal among those inducing $T$" doesn't "feel" like "the right condition". Let me argue that you can massage this into a "nicer" condition which still expresses the universal property of the EM construction in terms of adjunctions:

Claim: Let $\mathcal K$ be a 2-category. Then $\mathcal K$ admits Eilenberg-Moore objects if and only if the forgetful functor $Adj(\mathcal K) \to Mnd(\mathcal K)$ has a right adjoint.

Here $Mnd(\mathcal K)$ is defined as usual, and $Adj(\mathcal K)$ is the 2-category of adjunctions in $\mathcal K$, where a 2-cell is a 2-cell between the right adjoints.

Proof: As you say, Street proves one direction, so we prove the other. We have forgetful 2-functors

$$ Mnd(\mathcal K) \leftarrow Adj(\mathcal K) \xleftarrow{const} \mathcal K $$

By hypothesis, the leftmost of these has a right adjoint. We wish to show that the composite (sending $K \in \mathcal K$ to the identity monad on $K$) has a right adjoint (this characterization of EM objects is of course also due to Street in the same paper). It will suffice to show that the rightmost 2-functor has a right adjoint. And it does -- the right adjoint sends $(L : C {}^\to_\leftarrow D : R) \mapsto D$.


Notes:

  • If you think about the proof above, you'll see it also shows that this is true for a particular universal adjunction / EM object, even if these constructions aren't assumed to exist in general.

  • (The funny thing here is that the diagram above doesn't seem to be induced in an obvious way by somehow "homming" some diagram into $\mathcal K$. At least $Mnd(\mathcal K)$ is the lax Gray internal hom $[[ Mnd, \mathcal K]]$ -- or equally the lax functor category $Fun^{lax}(1, \mathcal K)$ -- but I don't know how to construct $Adj(\mathcal K)$ except "by hand". As a result, the verifications in the above also need to be done "by hand". But it works out!)

  • The condition that there be a terminal adjunctions inducing given monads $T$ says that the fiber of $Adj(\mathcal K) \to Mnd(\mathcal K)$ over $T$ has a terminal object. The condition that the right adjoint to $Adj(\mathcal K) \to Mnd(\mathcal K)$ exist is stronger in one sense. But it's also weaker, unless we have some way of knowing that this right adjoint should be fully faithful -- a question which Street's proof of the other direction of the Claim should shed some light on (er-- I suppose this follows representably from the fact that it's fully faithful in the case $\mathcal K = Cat$).

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  • $\begingroup$ Could you spell out why $\mathcal K$ admitting a terminal adjunction inducing $T$ for every monad $T$ is the same as admitting the specified right adjoint? Aren't the morphisms of the 2-category of adjunctions commutative squares, whereas the appropriate morphisms of "adjunctions inducing $T$" are commutative triangles? It's not obvious to me that the squares are forced to be trivial in this respect. $\endgroup$
    – varkor
    Commented Jul 25, 2022 at 16:29
  • $\begingroup$ @varkor Hmmm... Yeah, I think I've been too hasty, and answered a slightly different question. I've edited above to reflect this. $\endgroup$ Commented Jul 25, 2022 at 16:41
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    $\begingroup$ Thanks, I agree with the conclusion now. I think it is essentially the statement of Theoreme 4.4 of Auderset's Adjonctions et monades au niveau des 2-catégories. I don't necessarily disagree that the condition I'm asking for is generally not the most appropriate condition to ask for, but it does seem like a natural one to consider (at least from a historical perspective). $\endgroup$
    – varkor
    Commented Jul 25, 2022 at 16:49
  • $\begingroup$ @varkor I agree about the historical interest. And it sure would be fun to see some weird pathological counterexample! $\endgroup$ Commented Jul 25, 2022 at 16:54
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The answer is no: it is possible to have a terminal resolution without having an Eilenberg–Moore object.

Consider the 2-category $\mathbf{DagCat}$ of dagger categories, dagger functors, and natural transformations. A Frobenius monad is a monad in $\mathbf{DagCat}$ satisfying an additional law (see §5 of Heunen–Karvonen's Monads on dagger categories). According to §1 ibid., every Frobenius monad $T$ admits a terminal resolution whose apex $\mathbf{FEM}(T)$ is the category of Frobenius $T$-algebras (§6 ibid.). A Frobenius $T$-algebra is a $T$-algebra satisfying an additional law. Not every $T$-algebra is Frobenius: a counterexample is given in Example 6.4 ibid. However, postcomposition with the forgetful functor from $\mathbf{FEM}(T)$ always produces a Frobenius algbera by definition. Therefore, there exist $T$-algebras that do not factor uniquely through $\mathbf{FEM}(T)$, and consequently $\mathbf{FEM}(T)$ does not exhibit the Eilenberg–Moore object for $T$.

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