EDIT (3/89/14): ThereIt appears to beI made a minor errormistake in the Spitzweck papermy earlier version of this answer (and in a recent edit), so I've struck out that text abovedue to not fully understanding the notation in Markus Spitzweck's thesis. The proof of Corollary 8 says to look at Theorem 5. But Theorem 5 requires the operad to be cofibrantwhich I discuss above in the model category of sequences, i.estruck out text is not actually about strictly commutative monoids and thus has nothing to do with Tyler's answer. Spitzweck uses the notation $Comm_C$ to be projectively cofibrantdenote algebras over a particular (nowadays this would be called$E_\infty$ operad, and the $\Sigma$-cofibrancy for theof that operad assures us (by his Theorem 5) that algebras over the opeard have a semi-model structure. Unfortunately, $Com$ doesis not satisfy this$\Sigma$-cofibrant in general, so I don't believe Corollary 8 any more. In particularand in positive characteristic, I$CDGA(k)$ is not a semi-model category (I recently proved that if Corollary 8it were true then in fact CDGA($\mathbb{F}_p)$$k$) would be a (full) model category. Indeed, in any combinatorial situation you can get from a semi-model category to a model category, by Jeff Smith's recognition theorem. Spitzweck's Corollary 8 should have been a statement about the operad $E_\infty$, and then the result is related to the Mandell example mentioned above). Note that $Com$ satisfies the conditions of Theorem 5is $\Sigma$-cofibrant in $Ch(k)$ if $k$ is a field of characteristic zero, because in this setting all symmetric sequences are projectively cofibrant (I learned this from recent papers of John Harper). For non-zero characteristic it should be easy to construct examples which show that $Com$ is not projectively cofibrant as a symmetric sequence$\Sigma$-cofibrant.
Added another example of an almost model category, different from the previous examples
David White
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