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Jun 15, 2020 at 7:27 history edited CommunityBot
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Sep 29, 2016 at 11:35 comment added Bruno Stonek For a discussion of why "replace all maps with cofibrations" is not always the way to obtain a cofibrant diagram, see Strom, Modern Classical Homotopy Theory, section 6.4.2, paragraph "A common misconception". He defines what it means for a small category to be "tree-like", and proves that in this case it is true that a diagram is cofibrant iff all the arrows are cofibrant.
Feb 16, 2012 at 21:52 history edited David White CC BY-SA 3.0
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Jan 8, 2010 at 21:22 comment added Charles Rezk No. It's not even possible to "replace all arrows in the diagram with cofibrations" for most indexing categories. For instance, consider Delta^op, the small category which indexes simplicial objects. If F: Delta^op --> Top, is a functor which takes every arrow to a cofibration, then in particular it takes every arrow to an injection. But it turns out that any such functor must be constant (Delta^op is generated by arrows which have either left or right inverses, so such an F has to take all these to isomorphisms.)
Jan 8, 2010 at 20:41 comment added Joey Hirsh Would the statement be accurate if the diagram category if the diagram D satisfied: the nerve of D is contractible?
Jan 8, 2010 at 20:33 comment added Reid Barton Your first sentence is not accurate, except when the indexing category of the colimit is of a rather special type, such as in the case of pushouts. For instance, homotopy orbits for a discrete group G are a homotopy colimit indexed on the category with one object with automorphisms G, and the identity map of any object X is a cofibration, but the homotopy orbit space X_hG for the trivial action is not X in general (in Spaces, it's X x BG).
Jan 8, 2010 at 20:31 vote accept Joey Hirsh
Jan 8, 2010 at 20:30 answer added Charles Rezk timeline score: 10
Jan 8, 2010 at 20:25 answer added Reid Barton timeline score: 9
Jan 8, 2010 at 20:07 history asked Joey Hirsh CC BY-SA 2.5