Lemma 85.13.6. In Situation 85.3.3. If $K, K' \in D(\mathcal{C}_{total})$. Assume
$K$ is cartesian,
$\mathop{\mathrm{Hom}}\nolimits (K_ i[i - 1], K'_ i) = 0$ for $i > 1$.
Then any map $\{ K_ n \to K'_ n\} $ between the associated simplicial systems of $K$ and $K'$ comes from a map $K \to K'$ in $D(\mathcal{C}_{total})$.
Proof. Let $\{ K_ n \to K'_ n\} _{n \geq 0}$ be a morphism of simplicial systems of the derived category. Consider the objects $X_ n$ and Postnikov system $Y_ n$ associated to $K$ of Lemma 85.13.4. By (1) and Lemma 85.12.9 we have $X_ n = g_{n!}K_ n$. In particular, the map $K_0 \to K'_0$ induces a morphism $X_0 \to K'$. Since $\{ K_ n \to K'_ n\} $ is a morphism of systems, a computation (omitted) shows that the composition
is zero. As $Y_0 = X_0$ and as $Y_1$ fits into a distinguished triangle
we conclude that there exists a morphism $Y_1[1] \to K'$ whose composition with $X_0 = Y_0 \to Y_1[1]$ is the morphism $X_0 \to K'$ given above. Suppose given a map $Y_ n[n] \to K'$ for $n \geq 1$. From the distinguished triangle
we get an exact sequence
As $X_{n + 1}[n] = g_{n + 1!}K_{n + 1}[n]$ the last group is equal to
which is zero by assumption (2). By induction we get a system of maps $Y_ n[n] \to K'$ compatible with transition maps and reducing to the given map on $Y_0$. This produces a map
This map in any case has the property that the diagram
is commutative. Restricting to $\mathcal{C}_0$ we deduce that the map $\gamma _0 : K_0 \to K'_0$ is the same as the first map $K_0 \to K'_0$ of the morphism of simplicial systems. Since $K$ is cartesian, this easily gives that $\{ \gamma _ n\} $ is the map of simplicial systems we started out with. $\square$
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