MathOverflow Asked by Philippe Gaucher on August 26, 2020

I use the notation of this question. A non-decreasing continuous bijection from $[0,a]$ to $[0,b]$ where $a,bgeq 0$ are two real numbers is denoted by $[0,a] cong^+ [0,b]$. If $phi:[0,a]to U$ and $psi:[0,b]to U$ are two continuous maps for some topological space $U$ with $phi(a)=psi(0)$,

the map $phi*psi:[0,a+b]to U$ is the composition of the paths, $phi$ on $[0,a]$ and $psi$ on $[a,a+b]$. **All topological spaces are $Delta$-generated. Therefore all following categories are locally presentable.**

A **multipointed $d$-space** $X$ is a variant of Marco Grandis’ $d$-spaces. It consists of a topological space $|X|$, a subset $X^0$ (of states) of $|X|$ and a set of continuous maps (called execution paths) $mathbb{P}^{top}X$ from $[0,1]$ to $|X|$ satisfying the following axioms:

- for any $phiin mathbb{P}^{top}X$, $phi(0)$ and $phi(1)$ belong to $X^0$
- for any $phiin mathbb{P}^{top}X$, a composite $[0,1] cong^+ [0,1] stackrel{phi}longrightarrow |X|$ belongs to $mathbb{P}^{top}X$
- if $phi$ and $psi$ are two execution paths, all composites like

$[0,1] cong^+ [0,2] stackrel{phi*psi}longrightarrow |X|$ are execution paths.

Tu summarize, a multipointed $d$-space has not only a distinguished set of continuous paths but also a distinguished set of points (the other points are intuitively not interesting). Unlike Grandis’ notion, the constant paths are not necessarily execution paths. It is one of the role of the cofibrant replacement of the model category structure constructed in Homotopical interpretation of globular complex by multipointed d-space to remove from a multipointed $d$-space all points which do not belong to an execution path. The cofibrant replacement cleans up the underlying space by removing the useless topological structure.

It turns out that the model structure constructed in Homotopical interpretation of globular complex by multipointed d-space is the left determined model category with respect to the set of generating cofibrations $mathrm{Glob}(mathbf{S}^{n-1}) subset mathrm{Glob}(mathbf{D}^{n})$ for $ngeq 0$ and the map ${0,1} to {0}$ identifying two points where $mathbf{S}^{n-1}$ is the $(n-1)$-dimensional sphere, $mathbf{D}^{n}$ the $n$-dimensional disk, and where $mathrm{Glob}(Z)$ is the multipointed $d$-space whose definition is explained in the paper (I don’t think that it is important to recall it in this post).

Now here is the question. I would be interested in considering the multipointed $d$-spaces $vec{[0,1]^n}$ defined as follows

- The underlying space is the $n$-cube $[0,1]^n$
- The set of distinguished states is the set of vertices ${0,1}^n subset [0,1]^n$
- The set of execution paths is generated by the continuous maps from $[0,1]$ to $[0,1]^n$ such that of course $0$ and $1$ are mapped to a point of ${0,1}^n$ and such that these maps are nondecreasing with respect to each axis of coordinates.

The multipointed $d$-space $partialvec{[0,1]^n}$ is defined in the same way by removing the interior of the $n$-cube.

Using Vopenka’s principle and a result of Tholen and Rosicky, there exists a left determined model category structure with respect to the set of generating cofibrations $partialvec{[0,1]^n} subset vec{[0,1]^n}$ with $ngeq 0$ and $R:{0,1}to {0}$.

How is it possible to remove Vopenka’s principle from this statement ?

This question is probably too complicated for a post but if someone could give me a starting point, I would be very grateful. It is the reason why I ask the question anyway. Note: the presence of the map ${0,1}to {0}$ in the set of generating cofibrations is not mandatory because I start considering in other parts of my work model structures where I remove this map from the set of generating cofibrations.

~~ Note that category you've described is not locally presentable, but this is not a big deal -- if you use Delta-generated spaces (or some variant thereof) instead of general topological spaces, you're back in the locally presentable world.~~

**On getting by without Vopenka in general:**

In Tholen and Rosicky's paper they say (paragraph before Theorem 2.2) that Jeff Smith has claimed that left-determined model structures exist for locally presentable categories without Vopenka's principle. I'm not sure Smith has published his argument.

Existence of a left-determined model structure is essentially an "absolute" version of existence of Bousfield localizations. So some of the existing work on existence of Bousfield localizations under weaker hypotheses may be relevant.

**More to the point:**

But surely this misses the point, which is that you have a very specific category in front of you and it seems unlikely that the existence of this particular left-determined model structure really depends on set theory. In order to validate this argument, one needs a more detailed understanding of how the model structure works.

Unfortunately Olschok's theory doesn't apply since not every object is cofibrant, if I understand correctly.

It seems remarkable to me that in the related case you discuss, there is a reasonable description of a model structure which can be shown to be left-determined. I would think the way to proceed is to try to imitate whatever happens in that case.

**Use an adjunction:**

Is there a "geometric realization / nerve" adjunction between the old model category to the new one? I might suspect that if you projectively-induce the old left-determined model structure along such an adjunction, the result may be also left-determined, and in this case you would have a nice description of the model structure.

**A Reservation:**

I actually don't know whether the Quillen model structure on $mathsf{Top}$ is left-determined (the one on $mathsf{sSet}$, of course, is not). If it isn't, then my sense is it's unlikely that the left-determined model structure on multipointed $d$-spaces is actually what you want! I would suspect that the model structure you want is the projectively-induced model structure along some geometric realization adjunction with a more combinatorial category. Then if it turns out that the result is also left-determined, that's great, but if not, it's not a big deal.

Answered by Tim Campion on August 26, 2020

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