## cohomology – Verifying the quasi-isomorphism of two complexes

Let $$X$$ be a smooth variety over a number field $$k$$, with canonical morphism $$pi:X rightarrow mathrm{Spec} , k$$. Let $$mathcal{D}(k)$$ denote the derived category of bounded complexes of discrete $$Gamma_k (=mathrm{Gal}(bar{k}/k))$$-modules. We have the following truncated object in $$mathcal{D}(k)$$:

$$KD(X) = (tau_{leq 1}Rpi_*mathbb{G}_{m,X})(1).$$

This is a complex in degrees -1 and 0 and it is well-known that in our setting, it can simply be written as

$$(bar{k}(X)^* rightarrow mathrm{Div}(bar{X})).$$

By the canonical morphism $$i : mathbb{G}_{m,k} rightarrow tau_{leq 1}Rpi_*mathbb{G}_{m,X}$$, we define
$$KD'(X) = mathrm{Coker}(i)(1).$$

It is easy to check that $$KD'(X)$$ is quasi-isomorphic to the complex $$(bar{k}(X)^*/bar{k}^* rightarrow mathrm{Div}(bar{X})) in mathcal{D}(k).$$
However, I’ve seen that $$KD'(X)$$ can also be written as $$mathrm{Cone}(mathbb{G}_{m,k}(1) rightarrow KD(X))$$. How do we compute the cohomology groups of a mapping cone and verify that we do have a quasi-isomorphism?

## homological algebra – \$Omega^1_{B_bullet/A_bullet}\$ is acyclic if \$A_bullet to B_bullet\$ is quasi-isomorphism

Let $$A_bullet to B_bullet$$ be a quasi-isomorphism of simplicial rings in the sense of (P.62, I.3.1.7, Complexe Cotangent et Déformations I, Luc Illusie).

Then, we define the simplicial $$B_bullet$$-module of Kähler differentials $$Omega^1_{B_bullet/A_bullet}$$ by $$left(Omega^1_{B_bullet/A_bullet}right)_n := Omega^1_{B_n/A_n}$$ (P.119, ibid.).

My question is whether it follows that $$Omega^1_{B_bullet/A_bullet}$$ is acyclic (i.e. quasi-isomorphic to the zero module) from these assumptions.

The broader context is when I am trying to show that the cotangent complex is independent of the free resolution taken.

Suppose we are given an algebra $$R to S$$ which has two free simplicial resolutions $$P_bullet to Q_bullet$$ (i.e. they are quasi-isomorphic to $$S$$, and $$P_n$$ and $$Q_n$$ are free $$R$$-algebras).

Then, using these two simplicial resolutions, we define the cotangent complex $$L_{S/R}$$ to be $$Omega_{P_bullet/R} otimes_{P_bullet} S$$ and $$Omega_{Q_bullet/R} otimes_{Q_bullet} S$$, the former of which is isomorphic to $$Omega_{P_bullet/R} otimes_{P_bullet} Q_bullet otimes_{Q_bullet} S$$. My approach was to break down the question of whether $$Omega_{P_bullet/R} otimes_{P_bullet} S$$ and $$Omega_{Q_bullet/R} otimes_{Q_bullet} S$$ are quasi-isomorphic into two sub-questions:

1. Whether the map $$Omega_{P_bullet/R} otimes_{P_bullet} Q_bullet to Omega_{Q_bullet/R}$$ is a quasi-isomorphism.
2. Whether $$- otimes_{Q_bullet} S$$ sends quasi-isomorphisms of free $$Q_bullet$$-modules to quasi-isomorphisms.

For 1, I have fitted them in a short exact sequence (in general $$0 to$$ is not present, but for my case it is ok):
$$0 to Omega_{P_bullet/R} otimes_{P_bullet} Q_bullet to Omega_{Q_bullet/R} to Omega_{Q_bullet/P_bullet} to 0$$

So it suffices to show that $$Omega_{Q_bullet/P_bullet}$$ is acyclic, which is what I asked above.

For 2, I guess the usual proof with cones would go through ($$f$$ quasi-isomorphism $$implies$$ $$operatorname{cone}(f)$$ acyclic $$implies$$ $$operatorname{cone}(f) otimes_{Q_bullet} S = operatorname{cone}(f otimes_{Q_bullet} S)$$ acyclic $$implies$$ $$otimes_{Q_bullet} S$$ quasi-isomorphism), but I have not checked the details yet. I would appreciate if someone would give me a pointer on this, but I could also ask this in a later question.

## Show that \$hom_R(f, M)\$ is a quasi-isomorphism if \$f:P to P’\$ is a quasi-isomorphism of \$pi\$-projectives

For any complex of $$R$$-module M, how to show that $$hom_R(f, M)$$ is a quasi-isomorphism if $$f:P to P’$$ is a quasi-isomorphism of $$pi$$-projective complexes of $$R$$-modules.