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A useful result due to Linton is that for a cocomplete category $C$ and monad $T$ on $C$, if the category of algebras $C^T$ admits reflexive coequalisers, then it is cocomplete (see here for a sketch of a proof).

Are there any analogous results in the literature for enriched categories? More specifically, let $V$ be a monoidal category, let $C$ be a $V$-category admitting all weighted colimits, and let $T$ be a $V$-monad on $C$ admitting a $V$-category of algebras $C^T$. Is there a convenient collection of weights $\Phi$ such that, if $C^T$ admits $\Phi$-weighted colimits, then $C^T$ admits all weighted colimits? Ideally when $V = \mathrm{Set}$, this would reduce to Linton's characterisation. (Clearly if $C^T$ admits conical colimits and copowers, it is cocomplete, so I am interested in more minimal assumptions than these.) $V$ may be assumed sufficiently nice.

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  • $\begingroup$ For clarity, Fred Linton's result assumes that the functor preserves reflexive coequalisers and then makes clever use of their interaction with (infinitary) coproducts. $\endgroup$
    – Paul Taylor
    Commented Dec 16, 2022 at 15:54
  • $\begingroup$ @PaulTaylor: it is not necessary to assume that the underlying endofunctor preserves reflexive coequalisers, but when this is true (and $C$ is cocomplete), then in particular $C^T$ has reflexive coequalisers and so is cocomplete. $\endgroup$
    – varkor
    Commented Dec 16, 2022 at 15:56
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    $\begingroup$ Yes, you can assume that the reflexive coequalisers are already there, but I was trying to point out that you don't get them on the cheap. The literature claims that transfinite recursion is needed, but these seminar slides shows how they can be found as an example of my version of Pataraia's fixed point theorem. $\endgroup$
    – Paul Taylor
    Commented Dec 16, 2022 at 16:08
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    $\begingroup$ I suggest to check the nlab page about the "adjoint lifting theorem". There there is a "formal" treatment of how to propagate cocompleteness from the only existence of coequalizers of reflexive pairs. This might be inspiring for an enriched understanding. $\endgroup$
    – Ivan Di Liberti
    Commented Dec 17, 2022 at 6:03

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