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Let $W \subseteq \mathcal{C}$ be a wide subcategory of a category $\mathcal{C}$. The saturation $\overline{W}$ of $W$ is the class of maps of $\mathcal{C}$ which become isomorphisms in $\mathcal{C}[W^{-1}]$. Let $U \subseteq W$ be a wide subcategory of $W$ such that $\overline{U} = \overline{W}$. Then the obvious map between the hammock localizations $L^H(\mathcal{C},U) \to L^H(\mathcal{C},W)$ is a DK-equivalence.

Let's define a simplicial subcategory $L^H(\mathcal{C},U,W)$ of $L^H(\mathcal{C},W)$. It has all the objects of $L^H(\mathcal{C},W)$ and a simplex of $\mathrm{Hom}_{L^H(\mathcal{C},W)}(X,Y)$ belongs to $\mathrm{Hom}_{L^H(\mathcal{C},U,W)}(X,Y)$ if its vertices belong to $\mathrm{Hom}_{L^H(\mathcal{C},U)}(X,Y)$. That is, simplices in $\mathrm{Hom}_{L^H(\mathcal{C},U,W)}(X,Y)$ are hammocks in which vertical maps belong to $W$ and horizontal maps which go to the left belong to $U$. Then the map $L^H(\mathcal{C},U) \to L^H(\mathcal{C},W)$ factors through $L^H(\mathcal{C},U,W)$.

Question: What are sufficient and necessary conditions on $U$ and $W$ for the map $L^H(\mathcal{C},U,W) \to L^H(\mathcal{C},W)$ to be a DK-equivalence?

One such sufficient condition is that $W$ satisfies the 2-out-of-3 condition and every map in $W$ functorially factors as a section and a map in $U$ (of course, the dual condition is also sufficient). It is possible that the functoriality condition can be omitted, but the proof would be much harder. I don't think that the map in the question is always a DK-equivalence, so I also would like to see an example when it is not (it would be nice if $W = \overline{W}$ in this example).

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  • $\begingroup$ I really like this question. Just now I realized that when C is a model category (or, more generally, admits a three-arrow calculus), then U can be the union of the trivial fibrations and the trivial cofibrations, and you get the same result for any W, by 2 out of 3. Also, functoriality can be omitted thanks to arxiv.org/pdf/1001.4536.pdf. Not sure if these observations are enough of an answer, though. $\endgroup$ Commented Mar 14 at 21:33

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