Contractibility of the category of cosimplicial resolutions

Let γ:CM be a functor and define a cosimplicial resoultion of γ as a functor Γ:CMΔ such that

  • ΓC is Reedy cofibrant for every CC
  • for every C there is a natural weak equivalence w(C):ΓCcγC

We can define a category R=coRes(γ) where the morphisms are natural transformations η:Γ1Γ2 such that for all C the obvious triangles commute i.e. we have w2(C)ηC=w1(C) for all C.

I would like to understand why this category, as is well known, is
contractible.

Since I do not understand anything of the proof I found in the text I consulted, I am trying to prove it by myself in the following way:

  • A resolution exists because for every C, we can find a cofibrant object XC in MΔ and a weak equivalence XCcγC and this defines a functor X(C)=XC by functorial factorization.
  • For every ΓR, by functorial factoriazion there is a morphism XΓ.
  • If I call weak equivalence in R a map η such that ηC is a weak equivalence in the Reedy model structure in MΔ for all C, then given any map of resolutions η:Γ1Γ2, by commutativity of the triangle we have that η is a weak equivalence under this defintion.
  • Now, my naive intuition is that the contractibility of R should follow from the fact that if we formally invert all morphisms in R=coRes(γ), the resulting localization R[R1] is a simply connected groupoid, hence contractible.
  • I put on R the equivalence relation given by identifying all parallel morphisms, which is a congruence. In this way, all morphisms become invertible in the quotient so that I can call R/=R[R1] and I have the quotient functor q:RR[R1].
  • For every Γ, the arrow category Γq is contractible having initial object, so I conclude by Quillen’s theorem A.

Is this proof reasonable?

Edit The last bullet point is wrong because when I pass to the comma category I lose the initial object.

Also, apparently we cannot just pass to the quotient without using some extra propery of R: if it were possible to apply the reasoning I wanted to make, it would imply that any category with an object X such that Hom(X,A) and Hom(A,X) for all A would become contractible. And I just found counterexamples to this fact in this other question.

I still wonder if by using some more property of R, for example the fact that the maps I am inverting were all weak equivalences in some model structure, we can still deduce the contractibility of R from that of R[R1] along the quotient functor in this case.

Answer

Since you have functorial factorisations you should exploit that to the hilt.

If M is a model category with functorial factorisations then the category cM of cosimplicial objects in M, with the Reedy model structure, is also a model category with functorial factorisations. There is an obvious fully faithful embedding McM, so we may as well just forget about cosimplicial objects and just prove the following claim:

For every model category M with functorial factorisations and every diagram F:CM, the full subcategory Q(F) of the over-category [C,M]/F spanned by the componentwise cofibrant replacements of F is contractible.

Indeed, let Q:MM be a functor and let p:QidM be a natural transformation such that, for every object M in M, QM is a cofibrant object in M and pM:QMM is a weak equivalence in M. Such Q and p exist because M has functorial factorisations. Then, for every natural transformation α:FF and every object C in C, we have the following commutative square in M:
\require{AMScd}
\begin{CD}
Q F’ C @>{p_{F’ C}}>> F’ C \\
@V{Q \alpha_C}VV @VV{\alpha_C}V \\
Q F C @>>{p_{F C}}> F C
\end{CD}

This is all natural in C, so we actually have a commutative square in [\mathcal{C}, \mathcal{M}], hence a zigzag (Q F, p F) \leftarrow (Q F’, \alpha \bullet p F’) \rightarrow (F’, \alpha) in the overcategory [\mathcal{C}, \mathcal{M}]_{/ F}. But (Q F, p F) is a componentwise cofibrant replacement of F, and this is natural in F’, so we have a zigzag of natural transformations connecting the identity functor on \mathcal{Q} (F) and a constant functor. Therefore \mathcal{Q} (F) is contractible.

If you are geometrically inclined, you may think of the above proof as constructing a deformation retract of \mathcal{Q} (F) to a point. Of course, any space with a deformation retract to a point is contractible. The gist of the argument is widely applicable and can be used in contexts where one does not have a model structure per se – this, I think, is the point of Part II of Homotopy limit functors on model categories and homotopical categories.

Attribution
Source : Link , Question Author : giuseppe , Answer Author : Zhen Lin

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