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Jun 15, 2021 at 11:57 comment added Maxime Ramzi @ZhenLin : ah, so my guess would be wrong ! I think it would be lovely if you (or someone else) could find that "somewhere" to correct my answer - but to complete it one would need a proof that there is no alternative definition that can make it simplicial (I must confess I haven't read Dugger's paper in detail so I don't know if there's a uniqueness statement that would forbid this)
Jun 15, 2021 at 11:24 comment added Zhen Lin The functor $U \mathcal{C} \to \mathcal{M}$ is defined by taking a cosimplicial resolution (= componentwise Reedy-cofibrant replacement) of $\gamma : \mathcal{C} \to \mathcal{M}$ and then applying unenriched left Kan extension. I remember reading somewhere – one of Dugger's papers, I'm sure – that this is not necessarily simplicially enriched. But if it happens that $\gamma (-) \otimes \Delta^\bullet$ is a cosimplicial resolution of $\gamma$ then I think you do get a simplicially enriched functor.
Jun 15, 2021 at 10:32 history answered Maxime Ramzi CC BY-SA 4.0