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Nov 20, 2021 at 5:07 comment added Tim Campion I can think of precious few situations in homotopy theory where one classifies all diagrams weakly equivalent to a given diagram whose homotopy colimit agrees with their colimit. Would you be happy to understand some fairly general, sufficient conditions?
Nov 18, 2021 at 19:30 comment added Philippe Gaucher @DylanWilson The question is badly formulated and badly abstracted from my situation. Consider a coend $\int^{i} X(i)\times D(i)$. What condition should satisfy the diagram $D$ so that by replacing $X$ by a weakly equivalent diagram $Y$, one obtains a weakly homotopy equivalent coend ? (this coend does not have to calculate the homotopy colimit of $X$). I think that something like Theorem 11.5.1 should help.
Nov 18, 2021 at 13:44 comment added Dylan Wilson @DavidWhite That seems to only cover one case of D (the example in the question), unless I'm misunderstanding. Something like 11.5.1 seems closer, but there's something to check about the difference between 'cofibrant' diagrams (which are kinda like 'projective resolutions') and 'flat' diagrams (like flat resolutions).
Nov 18, 2021 at 7:46 comment added Philippe Gaucher @DylanWilson I asked the question because I suspect that a coend I have behaves (at least sometimes) as a homotopy colimit and the diagram $D$ does not look at all like the diagram in the question, nor I can see how to realize it as a simplicial set.
Nov 18, 2021 at 1:23 comment added David White Can this be deduced from Theorem 8.2.1 of Riehl's Categorical Homotopy Theory book?
Nov 17, 2021 at 20:51 comment added Dylan Wilson One guess could be something like: those D such that the augmentation to a point is a pointwise equivalence + such that the coend you wrote down is invariant under weak equivalences in the variable X. (i.e. something like "flat resolutions of the constant diagram at a point"). Possible proof: Induct over a 'cell' decomposition of X and use that Top is proper.
Nov 17, 2021 at 19:52 history asked Philippe Gaucher CC BY-SA 4.0