Let γ:C→M be a functor and define a cosimplicial resoultion of γ as a functor Γ:C→MΔ such that

- ΓC is Reedy cofibrant for every C∈C
- for every C there is a natural weak equivalence w(C):ΓC∼→c∗γC
We can define a category R=coRes(γ) where the morphisms are natural transformations η:Γ1→Γ2 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 XC∼→c∗γC and this defines a functor X(C)=XC by functorial factorization.
- For every Γ∈R, by functorial factoriazion there is a morphism X→Γ.
- 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:Γ1→Γ2, 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[R−1] 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[R−1] and I have the quotient functor q:R→R[R−1].
- For every Γ, the arrow category Γ↓q is contractible having initial object, so I conclude by Quillen’s theorem A.
Is this proof reasonable?

EditThe 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[R−1] 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 M→cM, 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:C→M, 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:M→M be a functor and let p:Q⇒idM be a natural transformation such that, for every object M in M, QM is a cofibrant object in M and pM:QM→M is a weak equivalence in M. Such Q and p exist because M has functorial factorisations. Then, for every natural transformation α:F′⇒F 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*