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Feb 19, 2013 at 6:54 comment added Karol Szumiło On the other hand it seems to me that the generating (acyclic) cofibrations in $\mathcal{K}$ are closed embeddings and that closed embeddings are preserved under pushouts and sequential colimits in $\mathcal{K}$. Hovey states this for $\mathcal{T}$ but implies that this is problematic for $\mathcal{K}$ but I don't see why it should fail.
Feb 19, 2013 at 6:53 comment added Karol Szumiło But again this doesn't seem useful since the mixed model structure is (as far as I understand) not known to be cofibrantly generated. In fact Hovey's remark on page 7 that you mention seems to say something similar about $\mathcal{K}$, namely that we don't have enough smallness properties to construct a model structure on the category of monoids.
Feb 19, 2013 at 6:47 comment added Karol Szumiło It seems to me that for the mixed model structure it is also rather simple. The fibrations of the mixed model structure are the same as those of the Strøm model structure. Thus the acyclic cofibrations of the mixed model structure are the same as those of the Strøm model structure i.e. acyclic Hurewicz cofibrations. Therefore the monoid axiom holds for the mixed model structure since it holds for the Strøm model structure.
Feb 18, 2013 at 18:11 vote accept David White
Feb 18, 2013 at 18:11 comment added David White @Karol: thanks for your comments, they are very helpful. My version of the Hovey preprint seems different from the one you found. Mine came from his website, and I'm guessing yours came from arxiv.org/abs/math/9803002. Anyway, yours seems to be the more up-to-date version, and it even comments at the bottom of page 7 that the monoid axiom holds for k-spaces, due to Lemma 2.3. So my last question then is about the mixed model structure. Do you have any ideas about whether or not the monoid axiom holds there? I'll think about it this week, but if you see the answer I'd love to hear it
Feb 18, 2013 at 8:50 comment added Karol Szumiło By the way, for the Strøm model structure this doesn't seem to be a useful observation since the construction of a model structure on the category of monoids (or modules over a monoid) uses the assumption that the original model category was cofibrantly generated.
Feb 16, 2013 at 13:31 comment added Karol Szumiło This of course applies to the Quillen model structure. For the Strøm model structure there is nothing to check since every object is cofibrant so crossing with any space preserves all acyclic cofibrations.
Feb 16, 2013 at 13:29 comment added Karol Szumiło I can't seem to locate a version of this preprint which contains the lemma you are talking about so I can't comment on its proof. (In the version I've found 1.5 is the definition of smallness.) When I say compact I mean "compact Hausdorff" in the most classical topological sense. Homming out of compact Hausdorff spaces commutes with sequential colimits of open embeddings. In the last paragraph I reduce my colimit to a colimit of open embeddings and in the last line I apply this to spheres in order to see that $X_0 \to \mathrm{Tel}_{\beta < \alpha} X_\beta$ is a $\pi_*$-isomorphism.
Feb 15, 2013 at 21:33 comment added David White Can you explain the last line a little bit more? What is the compact object? It seems like it must be $X_0$ for your proof to make sense, but I don't see why that should be compact. Also, this proof seems to only work for the Quillen model structure. Is there any chance it could also be made to work for the mixed model structure or the Strom?
Feb 15, 2013 at 20:46 comment added David White I'm using the terminology from Hovey's book and preprint, so K-space means compactly open subsets are open, and compactly generated means a K-space which is also weak Hausdorff. Lemma 1.5 from the preprint, shows the category T of compactly generated spaces (in Hovey's language) has the monoid axiom. He proves this using the fact that the generating trivial cofibrations $I^n$×0→$I^n$×I are closed inclusions of strong def retracts. His proof breaks down in K (ie K-spaces) because compact spaces are not finite (categorically). Does your proof secretly use compact $\Rightarrow$ finite anywhere?
Feb 13, 2013 at 7:59 history edited Karol Szumiło CC BY-SA 3.0
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Feb 12, 2013 at 20:28 history answered Karol Szumiło CC BY-SA 3.0