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Consider the class of cofibrations of the Quillen model structure, restricted to delta-generated topological spaces (the full subcategory of topological spaces generated by the colimits of simplices). Under Vopenka's principle, a left determined model structure w.r.t. a given class of cofibrations always exists if the underlying category is locally presentable. Since delta-generated spaces are locally presentable, how could this left determined model structure w.r.t. Quillen cofibrations look like ? Any idea ?

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Maybe I should say that I asked this question because I am trying to understand what the homotopy exchange property means for topological spaces. This notion is introduced by Marc Olschok in his PhD to generalize Lafont-Metayer-Worytkiewicz's construction of the folk model structures on globular $\omega$-categories. Marc told me that the homotopy exchange property is satisfied by the cylinder of topological spaces. So unless I am missing something, that implies that the Quillen model structure on delta-generated spaces is left determined, which is very weird. –  Philippe Gaucher Jun 22 '12 at 18:00

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I can answer my question now... Not only the Quillen model structure on $\Delta$-generated spaces is left determined, but also the hypothesis $\Delta$-generated can be removed. The left determined model structure exists by Marc Olschok's PhD. The Quillen model structure has the same class of cofibrations and more weak equivalences. So the left determined model structure has more fibrant objects, that is all topological spaces. So the left determined model structure and the Quillen model structure have the same class of cofibrations and the same class of fibrant objects (all topological spaces). Therefore they are equal.

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Thanks for posting the answer. It's nice to get these things resolved, even when the OP has to answer his own question. One question: when you say that the left determined model structure has more fibrant objects, you mean more than the Quillen model structure on $\Delta$-generated spaces, right? What do the fibrant objects there look like? –  David White Jun 28 '12 at 17:56
    
I mean: every trivial cofibration of the left determined one is a trivial cofibration of Quillen's, so dually, every fibrant of the Quillen's is fibrant in the left determined one, i.e. all spaces are fibrant in the left determined one. BTW, the same argument also proves that with Vopenka's principle, every combinatorial model category such that all objects are fibrant is left determined, fact that I had never noticed before asking this question. –  Philippe Gaucher Jun 28 '12 at 21:48
    
Actually, there is a slightly shorter argument: Marc Olschok already proves in his PhD that, in this case, all objects of the left determined one are fibrant. So the left determined one has the same cofibration and the same fibrant objects than Quillen's, so the two model structures are equal. –  Philippe Gaucher Jun 28 '12 at 21:53

Philippe's argument works well. I just wanted to remark that in this special case one does not need to invoke the homotopy exchange property explicitely as long as one already accepts that a model structure with the given cofibrations and fibrant object exists for the category $\mathrm{Top}_\Delta$ of $\Delta$-generated spaces.

Given a map $f\colon X\to Y$ in $\mathrm{Top}_\Delta$, first build the following diagram in $\mathrm{Top}$:

$$ \matrix{ X & \mathop{\longrightarrow}\limits^{\tilde{f}} & N_f & \mathop{\longrightarrow}\limits^{p'} & X \cr {\scriptstyle f} \big\downarrow {\ }& & {\ } \big\downarrow {\scriptstyle f'} & & {\ } \big\downarrow {\scriptstyle f} \cr Y & \mathop{\longrightarrow}\limits_t & Y^I & \mathop{\longrightarrow}\limits_{p_0} & Y \cr & & {\ } \big\downarrow {\scriptstyle p_1} & & \cr & & Y{\ } & & } $$

Here $(Y^I,p_0,p_1,t)$ is the usual path object on $Y$, $N_f$ is the pullback of $p_0$ and $f$, and $\tilde{f}$ is the map induced by $id_X$ and $f.t\colon X \to Y^I$. Observe that the $p_0$ and $p_1$ are trivial fibrations in $\mathrm{Top}$.

Define $\hat{f}\colon N_f\to Y$ as the composite $f'.p_0\colon N_f\to Y^I\to Y$. Then $f = \tilde{f}.\hat{f}\colon X\to N_f\to X$ is called the 'glueing factorization' of $f$.

The map $\hat{f}$ is always a fibratrion. This is the only appeal to classical algebraic topology.

Now apply the coreflection $k\colon \mathrm{Top}\to \mathrm{Top}_\Delta$ to that diagram.

Then we have:

(1) $k(\hat{f})$ is a fibration in $\mathrm{Top}_\Delta$ and the maps $k(p_0)$ and $k(p_1)$ are trivial fibrations in $\mathrm{Top}_\Delta$ because $k$ preserves fibrations and trivial fibrations.

(2) $k(p')$ is a trivial fibration in $\mathrm{Top}_\Delta$ because it is the pullback (in $\mathrm{Top}_\Delta$) of $k(p_0)$ along $k(f)$. Therefore $k(\tilde{f})$ lies in the smallest localizer by the 2-for-3-property.

Now suppose that $f$ is a weak equivalence in $\mathrm{Top}_\Delta$. Then the same holds for $k(f)$ and the 2-for-3 property yields that $k(f')$ is also a weak equivalence. Consequently $k(\hat{f}) = k(f').k(p_1)$ is a trivial fibration and $k(f) = k(\tilde{f}).k(\hat{f})$ is in the smallest localizer.

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Sorry, I have know idea why the rendering software suddenly trips over line (2) above. The text should read $k(p')$ is a trivial fibration in $\mathrm{Top}_\Delta$ because it is the pullback (in $\mathrm{Top}_\Delta$) of $k(p_0)$ along $k(f)$. Therefore $k(\tilde{f})$ lies in the smallest localizer by the 2-for-3-property. –  Marc Olschok Jul 6 '12 at 17:52
    
The trick was to put backticks on the outsides of the dollar signs for the part that was getting messed up –  David White Jul 6 '12 at 18:02

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