A question about a truncated object I was hoping someone could help me with the understanding of a particular truncated object. Here are some background:
For any object $A$ in an abelian category $\mathcal{A}$, we can view $A$ as an object in the category of complexes $\mathbf{C}(\mathcal{A})$ in $\mathcal{A}$ by setting $A$ as the degree zero object and the other terms in the complex as 0. Two complexes $A^\bullet$ and $B^\bullet$ in $\mathbf{C}(\mathcal{A})$ are quasi-isomorphic if the hypercohomology groups $H^i(A^\bullet)$ and $H^i(B^\bullet)$ are isomorphic for all $i$. By formally inverting all quasi-isomorphisms, we get the so-called derived category $\mathcal{D}(\mathcal{A})$ of $\mathcal{A}$.
In my setting, let $p : X \rightarrow Z$ be a faithfully flat morphism of finite type, where $Z$ is an integral regular Noetherian scheme and $X$ is a smooth geometrically integral scheme, both defined over a number field $k$. Thus we have a functor $p_*$ from the category of étale sheaves on $X$ to the category of étale sheaves on $Z$. Let $\mathcal{D}(Z)$ denote the category of bounded complexes of étale sheaves on $Z$ (same for $X$), this means that terms of sufficiently small or large degree simply vanish. Then we have derived functor $\mathbf{R}p_*: \mathcal{D}(X) \rightarrow \mathcal{D}(Z)$.
One obtains a complex $\mathbf{R}p_*\mathbb{G}_{m,X}$ in $\mathcal{D}(Z)$. As above, we write this complex as
$$\ldots \rightarrow 0 \rightarrow \mathbf{R}p_*\mathbb{G}_{m,X} \rightarrow 0 \rightarrow \ldots$$
where the sheaf is of degree zero and it is the only nonzero term in the complex.
Now we apply the truncation functor $\tau_{\leq 1}$, which sends a complex $A^\bullet$ to
$$\tau_{\leq 1}A^\bullet = \ldots \rightarrow A^{-2} \rightarrow A^{-1} \rightarrow A^0 \rightarrow \mathrm{ker}(d^1) \rightarrow 0 \rightarrow \ldots$$
and the shift functor $(-)[1]$ which sends $A^\bullet$ to the complex $A^\bullet[1]$ where the terms of degree $i$ in $A^\bullet$ are of degree $i-1$ in $A^\bullet[1]$.
Question. The truncated object $(\tau_{\leq 1}\mathbf{R}p_*\mathbb{G}_{m,X})[1]$ has trivial cohomology outside the degrees -1 and 0. But I don't see why it has nontrivial cohomology at 0 because this truncated object is simply a complex where the nonzero term is of degree -1 and everywhere else is 0. Is it possible that I understood the complex wrongly?
Context. This object was defined in the paper Descent theory for open varieties by Harari and Skorobogatov and it was denoted as $KD(X)$. This idea of using Galois hypercohomology was subsequently adopted by authors who attempt to generalise results about descent obstructions to the case for varieties $X$ where $\bar{k}[X]^* \neq \bar{k}^*$.
 A: One misconception in your post is in the definition of the derived category: we do not say that $A$ and $B$ are quasi-isomorphic if $H^i(A) \cong H^i(B)$ for all $i$. Instead we should define a quasi-isomorphism as a chain map inducing isomorphism on cohomology, and invert the quasi-isomorphisms, and this produces a richer theory. With the definition in your post every complex would be isomorphic to its cohomology in the derived category, which is absolutely not true in general: typically an object of the derived category carries much more information than its cohomology, and the whole point of the derived category is that we want to remember this extra information.
It seems also that you're misunderstanding how the derived functors work. To compute $Rp_\ast \mathbb G_m$ (say), you would pick your favorite injective resolution $\mathbb G_m \to I^\bullet$, or any $p_\ast$-acyclic resolution for that matter, and make the definition $Rp_\ast \mathbb G_m := p_\ast I^\bullet$. The magic of the derived category is that the result is well defined, i.e. there is a canonical isomorphism in the derived category between the outcome for any two choices of injective resolution.
In particular you really need to think of $Rp_\ast \mathbb G_m$ as a cochain complex. The classical derived functors $R^i p_\ast \mathbb G_m$ are the cohomologies of this cochain complex, but again, $Rp_\ast \mathbb G_m$ carries a lot more information than its cohomology. In the comments you wrote $Rp_\ast \mathbb G_m$ as a complex
$$ \ldots 0 \to R^0 p_\ast \mathbb G_m \to R^1 p_\ast \mathbb G_m \to R^2 p_\ast \mathbb G_m \to \ldots $$
(a cochain complex with all differentials zero) but only rarely will this be isomorphic to $Rp_\ast\mathbb G_m$.
