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Let $\mathcal{C}$ be a monoidal model category in the sense of Hovey's book. He assumes the following unit axiom not considered in other references (e.g. Schwede-Shipley): given a cofibrant replacement of the monoidal unit $q\colon QI\stackrel{\sim}\rightarrow I$ and a cofibrant object $X$, the morphisms $q\otimes X$ and $X\otimes q$ are weak equivalences.

This axiom obviously holds if $I$ is cofibrant. Moreover, if I'm not mistaken, a stronger version holds in all examples I know: $q\otimes X$ and $X\otimes q$ are weak equivalences for any $X$.

Does any of you know any example where this stronger version is false? Is it always true under some 'standard' assumptions on $\mathcal{C}$?

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    $\begingroup$ 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 $\endgroup$
    – David White
    Jan 10, 2012 at 16:05
  • $\begingroup$ 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 $\endgroup$
    – David White
    Jan 10, 2012 at 16:23
  • $\begingroup$ 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. $\endgroup$
    – David White
    Jan 10, 2012 at 18:48
  • $\begingroup$ The pushout product you say is just the identity in $QI\otimes X$ $\endgroup$
    – Fernando Muro
    Jan 10, 2012 at 19:58
  • $\begingroup$ 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 $\endgroup$
    – David White
    Jan 10, 2012 at 20:03

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I can only answer half of your question: namely, a standard condition under which the more general unit axiom holds. I don't know of any examples where Hovey's unit axiom holds but this more general one does not. The hypothesis is that cofibrant objects are flat, i.e. smashing with cofibrant objects preserves weak equivalences. This hypothesis comes up all the time in Hovey's work, and also appears in the paper of Schwede-Shipley where the Monoid Axiom is first introduced. There you need cofibrant objects to be flat in order to conclude that if $R\rightarrow S$ is a weak equivalence of ring objects, then $Ho(R-mod)\cong Ho(S-mod)$.

Suppose that cofibrant objects are flat. Then we know Hovey's unit axiom automatically, since $QI\rightarrow I$ is a weak equivalence and so for any cofibrant $Y$, $Y$ smashed with this map is still a weak equivalence. To see that the more general unit axiom holds, let $X$ be any object (not necessarily cofibrant). Then we have the following commutative diagram:

$$ \begin{array}{rrrr} QI\otimes QX & \rightarrow & QX & \rightarrow & X\\\ \downarrow & & & & \downarrow \\\ QI \otimes X & & \rightarrow & & X \end{array} $$

Here the top maps are weak equivalences by Hovey's unit axiom and by the definition of $QX$. The left vertical map is a weak equivalence because cofibrant objects are flat and $QI$ is cofibrant. The right vertical is clearly a weak equivalence because it's the identity. Thus, by the 2-out-of-3 property, the bottom horizontal is a weak equivalence.

EDIT: It should go without saying, but in a non-symmetric setting "cofibrant objects are flat" means both $K\otimes f$ and $f\otimes K$ are weak equivalences. So the argument above works if you apply the twist natural transformation to everything, and we also have $X\otimes QI \rightarrow X$ is a weak equivalence.

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    $\begingroup$ At this link you can find an example where cofibrant objects are not flat. My gut says that if you tweak the example to be $T$-alg, then it'll be a monoidal model category and Hovey's unit axiom will hold. Since cofibrant objects are not flat, maybe this is a natural place to look for a counter-example to your claim. books.google.com/… $\endgroup$
    – David White
    Jan 18, 2012 at 14:41

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