3
$\begingroup$

Consider a category $\mathcal C$ with a weak factorization system which is functorial. Let $*$ be an initial object. If $X\in \mathcal C$, denote by $\eta_X:*\to X$ the unique map. Using the given wfs, it factors as as $*\stackrel{\eta_{QX}}{\to} QX \stackrel{q_X}{\to} X $.

Consider the following assertion:

(A) The maps $Q\eta_X:Q*\to QX$ and $Q*\stackrel{q_*}{\to} * \stackrel{\eta_{QX}}{\to} QX$ coincide.

This is a minimalistic scenario: I'm interested in the situation where $\mathcal C$ is a (cofibrantly generated) model category with functorial factorizations and the wfs is (cofibrations, acyclic fibrations), so $Q$ is a cofibrant replacement functor. I would like to know the following:

Is it possible to modify the functorial factorization of the wfs (cofibrations, acyclic fibrations) of $\mathcal C$ without changing the model category structure (i.e. keeping the same classes of cofibrations, fibrations, weak equivalences) so that assertion (A) is true?

Manipulating some diagrams and the condition of functoriality lets us conclude that (A) is true if $q_X:QX\to X$ is a monomorphism, or if $\eta_{Q*}:*\to Q*$ is an epimorphism.

Is it possible to modify the functorial factorization of the wfs (cofibrations, acyclic fibrations) so that $\eta_{Q*}$ is an epimorphism? An isomorphism? The identity $id_*$?

We could be more ambitious with this last question: it would be nice if we could change the functorial factorization so that it factors any identity map through identity maps (rather than asking that it does this just with the identity map of $*$).

For background, I'd add that my category $\mathcal C$ is the category of commutative $\mathbb S$-algebras of Elmendorf, Kriz, Mandell, May (EKMM), so $*$ is $\mathbb S$.

Improve this question
$\endgroup$
9
  • $\begingroup$ @HarryGindi That's indeed the case! I'll edit it in, thank you. $\endgroup$
    – Bruno Stonek
    Nov 19, 2018 at 16:53
  • $\begingroup$ Ah, in this case is it not possible to just re-define $Q\ast$ to be $\ast$ itself, and leave the rest of the values of $Q$ unchanged? I think all of the relevant naturality squares will automatically commute, since $\ast$ is initial. $\endgroup$
    – Aaron Mazel-Gee
    Nov 19, 2018 at 17:40
  • $\begingroup$ @AaronMazel-Gee That's what I said in chat last week, but I wasn't sure enough to give it as an answer. It's like a cheeky extension by zero of the restriction to $C-\{*\}$ $\endgroup$
    – Harry Gindi
    Nov 19, 2018 at 17:55
  • $\begingroup$ @AaronMazel-Gee I don't see how this would work. For simplicity, suppose instead of modifying the whole functorial factorization we just modify the functor $Q$, so we ask ourselves whether we can construct $Q'$, an endofunctor of $\mathcal C$ such that $Q’X$ is $QX$ if $X\not= *$ and is $*$ if $X=*$. OK, then on an arrow $X\to Y$ it should be $QX\to QY$ if $X,Y\not= *$, it should be $*\to QY$ if $X=*$, and it should be $QX\to Q*\to *$ if $X\not=*$ and $Y=*$. But this is not always a functor. $\endgroup$
    – Bruno Stonek
    Nov 20, 2018 at 12:50
  • $\begingroup$ @HarryGindi (cont.) Indeed, consider a composition $X\stackrel{f}{\to} *\stackrel{g}{\to} Z$ where $X$ and $Z$ are not $*$. Then $Q’$ maps the composition to $QX\to Q*\to QZ$. On the other hand, the composition of the $Q’$’s of these arrows is $QX\to Q*\to *\to QZ$. So now the question is whether $Qg:Q*\to QZ$ is equal to $Q*\to *\to QZ$, which is a case of assertion (A) above, which doesn’t hold for $Q$ (if it did, there would be no need to go through all this). $\endgroup$
    – Bruno Stonek
    Nov 20, 2018 at 12:50

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.