Contractibility of the category of cosimplicial resolutions

Question

Let $\gamma : \mathcal{C} \to \mathcal{M}$ be a functor and define a cosimplicial resoultion of $\gamma$ as a functor $\Gamma: \mathcal{C} \to \mathcal{M}^{\Delta}$ such that

• $\Gamma C$ is Reedy cofibrant for every $C \in \mathcal{C}$
• for every $C$ there is a natural weak equivalence $w(C):\Gamma C \xrightarrow{\sim} c^* \gamma C $

We can define a category $\mathcal{R}=\text{coRes}(\gamma)$ where the morphisms are natural transformations $\eta:\Gamma_1 \to \Gamma_2$ such that for all $C$ the obvious triangles commute i.e. we have $w_2(C) \circ \eta_C = w_1(C)$ 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 $X_C$ in $\mathcal{M}^{\Delta}$ and a weak equivalence $X_C \xrightarrow{\sim} c^*\gamma C$ and this defines a functor $X(C)=X_C$ by functorial factorization.
• For every $\Gamma \in \mathcal{R},$ by functorial factoriazion there is a morphism $X \to \Gamma.$
• If I call weak equivalence in $\mathcal{R}$ a map $\eta$ such that $\eta_C$ is a weak equivalence in the Reedy model structure in $\mathcal{M}^{\Delta}$ for all $C,$ then given any map of resolutions $\eta:\Gamma_1 \to \Gamma_2,$ by commutativity of the triangle we have that $\eta$ is a weak equivalence under this defintion.
• Now, my naive intuition is that the contractibility of $\mathcal{R}$ should follow from the fact that if we formally invert all morphisms in $\mathcal{R}=\text{coRes}(\gamma)$, the resulting localization $\mathcal{R}[\mathcal{R}^{-1}]$ is a simply connected groupoid, hence contractible.
• I put on $\mathcal{R}$ 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 $\mathcal{R}/{\sim}=\mathcal{R}[\mathcal{R}^{-1}]$ and I have the quotient functor $q:\mathcal{R}\to \mathcal{R}[\mathcal{R}^{-1}].$
• For every $\Gamma,$ the arrow category $\Gamma \downarrow q$ 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 $\mathcal{R}$: if it were possible to apply the reasoning I wanted to make, it would imply that any category with an object $X$ such that $\text{Hom}(X,A) \neq \emptyset$ and $\text{Hom}(A,X) \neq \emptyset$ 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 $\mathcal{R}$, 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 $\mathcal{R}$ from that of $\mathcal{R}[\mathcal{R}^{-1}]$ along the quotient functor in this case.

Answer 1

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

If $\mathcal{M}$ is a model category with functorial factorisations then the category $\mathbf{c}\mathcal{M}$ of cosimplicial objects in $\mathcal{M}$, with the Reedy model structure, is also a model category with functorial factorisations. There is an obvious fully faithful embedding $\mathcal{M} \to \mathbf{c} \mathcal{M}$, so we may as well just forget about cosimplicial objects and just prove the following claim:

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

Indeed, let $Q : \mathcal{M} \to \mathcal{M}$ be a functor and let $p : Q \Rightarrow \textrm{id}_\mathcal{M}$ be a natural transformation such that, for every object $M$ in $\mathcal{M}$, $Q M$ is a cofibrant object in $\mathcal{M}$ and $p_M : Q M \to M$ is a weak equivalence in $\mathcal{M}$. Such $Q$ and $p$ exist because $\mathcal{M}$ has functorial factorisations. Then, for every natural transformation $\alpha : F' \Rightarrow F$ and every object $C$ in $\mathcal{C}$, we have the following commutative square in $\mathcal{M}$: $$\require{AMScd} \begin{CD} Q F' C @>{p_{F' C}}>> F' C \\ @V{Q \alpha_C}VV @VV{\alpha_C}V \\ Q F C @>>{p_{F C}}> F C \end{CD}$$ 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.