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.

  • $\begingroup$ Isn't this much easier? In coRes($\gamma$), every object comes equipped with a weak equivalence to the terminal object. So it's contractible. Sure, if you like, you could point out that the homotopy category (of the Reedy model structure, say) is a groupoid, by the two out of three property for weak equivalences, but I think the statement is just obvious. Unless there's a subtlety I'm missing? $\endgroup$ Commented Aug 26, 2020 at 17:07
  • $\begingroup$ Why is the object terminal? I don’t see why the morphism is unique, maybe it’s me $\endgroup$ Commented Aug 26, 2020 at 17:11
  • $\begingroup$ @DavidWhite If by terminal object you mean the identity $c^*\gamma \to c^*\gamma$, then I think this does not belong to $\text{coRes}\gamma$ since $c^*\gamma C$ is not necessarily cofibrant. If instead you mean the object I call $X,$ then I don't see why the morphism should be unique. But it's possible I made a big deal out of something obvious? $\endgroup$ Commented Aug 26, 2020 at 17:32
  • 1
    $\begingroup$ There is nothing in your last two bullet points that uses special properties of $\mathcal{R}$, so it must be wrong. $\endgroup$
    – Zhen Lin
    Commented Aug 27, 2020 at 1:03
  • 2
    $\begingroup$ It's called proving too much. Your argument, as it stood, would have shown that any connected category whatsoever is contractible, which is obviously nonsense, as you realised. It seems you have also realised your other mistake, which is that you have not shown $\mathcal{R} / {\sim}$ is isomorphic to $\mathcal{R} [\mathcal{R}^{-1}]$ or, equivalently, that $\mathcal{R} [\mathcal{R}^{-1}]$ is contractible. $\endgroup$
    – Zhen Lin
    Commented Aug 27, 2020 at 3:39

1 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.

  • $\begingroup$ Thanks but I am not interested in another proof, there's a proof of this result on any book and they are all similar to this one. I wanted to see if my proof could work and I realized there is a point that does not work (thank you for making me think more about the last bullet). I never asked for a different proof so I'm sorry but I don't see how this is an answer to my question. $\endgroup$ Commented Aug 27, 2020 at 3:12
  • $\begingroup$ I'm surprised you say that you "don't understand anything of the proof", then – this is a much easier proof than what you proposed. It doesn't even rely on Quillen's Theorem A, which in my view is a non-trivial result concerning homotopy colimits. $\endgroup$
    – Zhen Lin
    Commented Aug 27, 2020 at 3:47
  • $\begingroup$ OK. But how do you show that a zig-zag of natural transformations connecting the identity functor and a constant functor is a sufficient condition for contractibility? It seems that is where all the meat of the proof is. $\endgroup$ Commented Aug 31, 2020 at 0:54
  • $\begingroup$ I see the analogy in the sense that if we define a homotopy between functors as a zig-zag of natural transformations, then clearly by your proof we retract $Q(F)$ to a point by deformation. But why is this the right notion of homotopy which agrees with the definition of contractibility that uses the nerve? $\endgroup$ Commented Aug 31, 2020 at 1:00
  • $\begingroup$ Apply the nerve functor. The natural transformations literally become simplicial homotopies. If you prefer to work with actual topological spaces then apply the geometric realisation functor as well. $\endgroup$
    – Zhen Lin
    Commented Aug 31, 2020 at 1:05

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