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Jan 18, 2012 at 19:33 vote accept Fernando Muro
Jan 18, 2012 at 14:27 answer added David White timeline score: 5
Jan 10, 2012 at 20:03 comment added David White Oops, I meant to write $f:QI \rightarrow I$ and $0\rightarrow X$. And with this it's clear why the pushout product axiom doesn't imply the unit axiom, since the map $f$ is not a cofibration. Thanks for finding my error
Jan 10, 2012 at 19:58 comment added Fernando Muro The pushout product you say is just the identity in $QI\otimes X$
Jan 10, 2012 at 18:48 comment added David White I have a question. This map $q\otimes X$ is the pushout product of $f:0\hookrightarrow QI$ and $id_X$, right? So shouldn't the pushout product axiom say that for $X$ cofibrant this map must be a trivial cofibration (since $id_X$ is a trivial cofibration)? I guess what I'm asking is why we need the Unit Axiom at all. According to Schwede-Shipley's remark, it's to be sure $I$ represents the unit on the homotopy level, but it seems to me that it should be implied by the Pushout Product axiom.
Jan 10, 2012 at 16:23 comment added David White One more thing: my gut says there must be examples where the stronger version fails, but I haven't come up with any so far. I keep feeling this stronger version would imply some other property on $\mathcal{C}$. The comments to this old question of mine might help: mathoverflow.net/questions/73704
Jan 10, 2012 at 16:05 comment added David White Hi Fernando. If you're thinking about the same Schwede-Shipley paper as I am (Algebras and Modules in Monoidal Model Categories), then they do mention the Unit Axiom. In my version it's Remark 3.2, right before the definition of the Monoid Axiom. Still, this doesn't answer your question about the stronger statement
Jan 10, 2012 at 12:54 history asked Fernando Muro CC BY-SA 3.0