Impact
~23k
people reached
- 0 posts edited
- 0 helpful flags
- 106 votes cast
Sep
17 |
awarded | Yearling |
Jul
17 |
accepted | Stabilization of a generic pointed model category |
Jul
16 |
comment |
Stabilization of a generic pointed model category
P.S.: Of course, my comment about non-functoriality applies only when trying to apply the construction to non-cofibrantly generated model categories. |
Jul
15 |
comment |
Stabilization of a generic pointed model category
@DavidWhite: I'll go through the references you gave in the second to last paragraph. As to writing down the suspension functor via framings, at appears to me that one would need functorial factorizations to get functorial cosimplicial frames of cofibrant replacements. Without functoriality, I only see how to get a well-defined suspension functor on the homotopy category. So one may have to check how to extends Hovey's machinery to "functors, well-defined only up to homotopy". |
Jul
14 |
comment |
Stabilization of a generic pointed model category
@White: Thanks. So my question remains. The question is less about a particular application than about getting a complete picture of what is going on. |
Jul
13 |
comment |
Stabilization of a generic pointed model category
Is every model category Quillen equivalent to a left proper combinatorial one? |
Jul
13 |
revised |
Stabilization of a generic pointed model category
reference added |
Jul
13 |
asked | Stabilization of a generic pointed model category |
May
9 |
awarded | Popular Question |
Feb
4 |
comment |
When are homotopy categories of model categories closed modules over the homotopy category of $(\infty, 1)$-categories?
Insofar as every (left proper, combinatorial) model category is Quillen equivalent to a simplicial one, I would compare it to the fact that, for example, every monoidal category is equivalent to a strict one. So one could call a Quillen equivalent simplicial model category a strictification of the original model category. As in the case of monoidal categories, I think it also makes sense to study the non-strict versions (or presentations). So I can rephrase my question as follows: What is a good property that makes a model category Quillen equivalent to one enriched over $\mathrm{Joyal}$? |
Feb
4 |
comment |
When are homotopy categories of model categories closed modules over the homotopy category of $(\infty, 1)$-categories?
This is too strong for being what I would call a natural condition. Every homotopy category of a simplicial model category (one that is enriched over $\mathrm{Quillen}$) can easily be seen to be a closed module over $h\mathrm{Quillen}$. But the surprising (?) fact is that every (simplicial or not) model category's homotopy category is a closed module over $h\mathrm{Quillen}$. |
Feb
4 |
comment |
When are homotopy categories of model categories closed modules over the homotopy category of $(\infty, 1)$-categories?
Could you give me link to Hovey's definition of a Joyal-model category? |
Feb
4 |
comment |
When are homotopy categories of model categories closed modules over the homotopy category of $(\infty, 1)$-categories?
@Muro I didn't spell it out. Added it. |
Feb
4 |
revised |
When are homotopy categories of model categories closed modules over the homotopy category of $(\infty, 1)$-categories?
added 149 characters in body |
Feb
4 |
comment |
When are homotopy categories of model categories closed modules over the homotopy category of $(\infty, 1)$-categories?
@Muro You won't get the natural closed module structure over $h\mathrm{Joyal}$ in the case of $\mathrm{Joyal}$. |
Feb
4 |
asked | When are homotopy categories of model categories closed modules over the homotopy category of $(\infty, 1)$-categories? |
Dec
13 |
awarded | Nice Question |
Jul
2 |
awarded | Curious |
Dec
27 |
accepted | Is the ideal membership problem solvable for differential ideals? Is there a good notion of a Gröbner basis? |
Dec
26 |
comment |
Is the ideal membership problem solvable for differential ideals? Is there a good notion of a Gröbner basis?
The ring $\Omega$ is super-commutative; the generators $x_i$ are of degree $0$, the generators $dx_i$ are of degree $1$. There is just one differential operator, the derivative $d$ of degree $1$, which fulfills $d \circ d = 0$, that is $d(dx_i) = 0$. |