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Jan 18 at 11:41 comment added Tom Goodwillie @Tyrone: Yes, definitely.
Jan 18 at 5:16 comment added Tyrone I feel that Tim's complete answer should be the accepted answer. I hope this is understandable.
Jan 16 at 22:26 history edited Tom Goodwillie CC BY-SA 4.0
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Jan 16 at 21:17 comment added Tim Campion Ok, I’ve added my observations as another answer. I think I’ve closed all the loopholes now.
Jan 16 at 18:26 comment added Tyrone Try this: acyclic fibrations must be surjective (otherwise there aren't enough cofibrant objects). Thus $\Delta^0$ is cofibrant. On the other hand, $\bigsqcup\Delta^0\rightarrow\bigsqcup\Delta^1$ would be a weak equivalence if both $\Delta^0,\Delta^1$ were cofibrant. Thus $\Delta^1$ is not cofibrant. Thus $\Delta^0\rightarrow\Delta^1$ is not a cofibration.
Jan 16 at 17:31 history edited Tom Goodwillie CC BY-SA 4.0
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Jan 16 at 16:29 comment added Tom Goodwillie @Tim Campion: Going back to your first comment, if the map $X\to \Delta^1$ is surjective, that gives you a right inverse, but the right inverse might not be compatible with the maps from $\Delta^0$.
Jan 16 at 14:56 comment added Tom Goodwillie To say that a morphism $f$ is either a cofibration or a fibration in any model structure is to say that for every weak factorization system $f$ is in either the left class or the right class. FWIW, in $Set$ this means precisely that $f$ is either injective or surjective.
Jan 16 at 13:25 vote accept Tyrone
Jan 18 at 5:16
Jan 16 at 13:25 comment added Tyrone I really like this! It's exactly the kind of argument I was hoping to see. I was waiting to accept in case David wanted to address the comments under his answer. Thanks also for your input, Tim!
Jan 16 at 13:09 history edited Tom Goodwillie CC BY-SA 4.0
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Jan 16 at 13:07 comment added Tom Goodwillie Oh, good! This completes the argument.
Jan 15 at 12:38 history edited Tom Goodwillie CC BY-SA 4.0
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Jan 15 at 12:20 history answered Tom Goodwillie CC BY-SA 4.0