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.