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1. • CommentRowNumber1.
   • CommentAuthorUrs
   • CommentTimeMar 12th 2010
   • (edited Mar 12th 2010)
   • PermaLink
   Author: Urs
   Format: Markdownexpanded [[(infinity,1)-category of (infinity,1)-functors]] : more details, more statements, more proofs. In particular concerning the model by the [[global model structure on functors]].

   expanded (infinity,1)-category of (infinity,1)-functors : more details, more statements, more proofs. In particular concerning the model by the global model structure on functors.

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeMar 15th 2010
   • PermaLink
   Author: Urs
   Format: Markdownadded remakrs and pointers from [[global model structure on functors]] to [[(infinity,1)-category of (infinity,1)-functors]]. What I still don't fully know is the general statement what invariant notion precisely the Reedy model structures model.

   added remakrs and pointers from global model structure on functors to (infinity,1)-category of (infinity,1)-functors.

   What I still don't fully know is the general statement what invariant notion precisely the Reedy model structures model.

3. • CommentRowNumber3.
   • CommentAuthorMike Shulman
   • CommentTimeMar 15th 2010
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   Author: Mike Shulman
   Format: MarkdownSince the weak equivalences in the Reedy model structure are the same as those in the projective and injective model structures, it should model the same <latex>(\infty,1)</latex>-category, no?

   Since the weak equivalences in the Reedy model structure are the same as those in the projective and injective model structures, it should model the same -category, no?

4. • CommentRowNumber4.
   • CommentAuthorUrs
   • CommentTimeMar 15th 2010
   • PermaLink
   Author: Urs
   Format: MarkdownRight. But the Reedy model structure exists without assumptions on the codomain. What if the codomain is not combinatorial, for instance? Does the Reedy model structure still present the (oo,1)-functor (oo,1)-category? In any case, I see that some polishing of the discussion of the relation to the proj/in j-structure at [[Reedy model structure]] was in order: created section <a href="http://ncatlab.org/nlab/show/Reedy+model+structure#Properties">Properties</a>.

   Right. But the Reedy model structure exists without assumptions on the codomain. What if the codomain is not combinatorial, for instance? Does the Reedy model structure still present the (oo,1)-functor (oo,1)-category?

   In any case, I see that some polishing of the discussion of the relation to the proj/in j-structure at Reedy model structure was in order: created section Properties.

5. • CommentRowNumber5.
   • CommentAuthorMike Shulman
   • CommentTimeMar 16th 2010
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   Author: Mike Shulman
   Format: MarkdownGood question. I would be surprised if the answer were no, but I don't actually know. Note that the projective model structure requires the codomain only to be cofibrantly generated. There do, of course, exist model categories that aren't cofibrantly generated.

   Good question. I would be surprised if the answer were no, but I don't actually know. Note that the projective model structure requires the codomain only to be cofibrantly generated. There do, of course, exist model categories that aren't cofibrantly generated.