Consider the class of cofibrations of the Quillen model structure, restricted to deltagenerated 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 deltagenerated spaces are locally presentable, how could this left determined model structure w.r.t. Quillen cofibrations look like ? Any idea ?

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. 


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 2for3property. Now suppose that $f$ is a weak equivalence in $\mathrm{Top}_\Delta$. Then the same holds for $k(f)$ and the 2for3 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. 

