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Aug 20, 2019 at 21:07 comment added Ben Wieland The point of the original question is that a model does not support every morphism. Tyler's example of a functor from $B\mathbb N$ is just the choice of an object and a morphism. If you allow Kan complexes or finite CW complexes, then they are cofibrant-fibrant and thus they support arbitrary morphisms. In the purely homotopical setting, the object supports the morphism. Tyler's example was $S^1$, which is as nicely cellular as you can imagine: represent it as the constant simplicial object. Maybe having to spell out the cells allows counterexamples, but not that one.
Aug 20, 2019 at 4:44 comment added Gregory Arone @BenWieland I am feeling a little slow. Why it is not a counterexample?
Aug 19, 2019 at 23:24 comment added Tyler Lawson @BenWieland yes, absolutely. you can replace each object in a diagram with any weakly equivalent object and get an equivalent diagram at the cost of working coherently.
Aug 19, 2019 at 21:28 comment added Ben Wieland If the "more general" version is purely homotopical, Tyler's example isn't a counterexample, is it? (a counterexample to an even more general version) And a nitpick, just a problem for objects, if a "finite homotopy colimit" includes retracts, then you run into Wall finiteness problems, but @R.vanDobbendeBruyn's version doesn't.
Aug 19, 2019 at 17:32 comment added R. van Dobben de Bruyn A possible more general finiteness condition could be the assumption that every object in the diagram has finitely many maps out of it. (This has popped up in some inductive constructions I ran into; examples include finite diagrams and semisimplicial objects.)
Aug 19, 2019 at 17:23 history edited Gregory Arone CC BY-SA 4.0
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Aug 19, 2019 at 16:39 history edited Gregory Arone CC BY-SA 4.0
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Aug 19, 2019 at 16:32 history edited Gregory Arone CC BY-SA 4.0
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Aug 19, 2019 at 16:28 comment added Tyler Lawson @AchimKrause That is, indeed, a great deal simpler than what I had in mind.
Aug 19, 2019 at 16:20 comment added Achim Krause In Tyler's example, can't we just observe that for any finite simplicial set $K$, only finitely many maps on $\pi_1$ are induced by endomorphisms $K\to K$? This is because there are only finitely many maps $K\to K$ in total.
Aug 19, 2019 at 15:17 history edited Gregory Arone CC BY-SA 4.0
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Aug 19, 2019 at 15:12 comment added Gregory Arone Hi Tyler, thanks. This sounds right and suggests I should impose a local finiteness condition on the indexing diagram (such as, there can be only finitely many morphisms between any two objects). What I really wanted to know was about simplicial objects: can any simplicial object in finite CW complexes be approximated by a simplicial object in finite simplicial sets? I will edit the question.
Aug 19, 2019 at 15:10 history edited Gregory Arone CC BY-SA 4.0
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Aug 19, 2019 at 14:07 comment added Tyler Lawson Hi Greg, I suspect that the map $B\Bbb N \to CW$, representing $S^1$ with the endomorphisms $2^k$ for $k > 0$, can't be realized by a diagram of finite simplicial sets, but to prove it I'd want to come up with some explicit growth condition on the number of edges needed to represent elements of $\pi_1$ for a fixed finite model. This may be a little harder for s.sets than for $\infty$-categories because the mapping complexes aren't Kan complexes.
Aug 19, 2019 at 9:15 history edited Gregory Arone CC BY-SA 4.0
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Aug 19, 2019 at 8:08 history edited Gregory Arone CC BY-SA 4.0
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Aug 19, 2019 at 7:58 history edited Gregory Arone CC BY-SA 4.0
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Aug 19, 2019 at 7:48 history asked Gregory Arone CC BY-SA 4.0