Lemma 85.13.5. In Situation 85.3.3. If $K, K' \in D(\mathcal{C}_{total})$. Assume
$K$ is cartesian,
$\mathop{\mathrm{Hom}}\nolimits (K_ i[i], K'_ i) = 0$ for $i > 0$, and
$\mathop{\mathrm{Hom}}\nolimits (K_ i[i + 1], K'_ i) = 0$ for $i \geq 0$.
Then any map $K \to K'$ which induces the zero map $K_0 \to K'_0$ is zero.
Proof. Consider the objects $X_ n$ and the Postnikov system $Y_ n$ associated to $K$ in Lemma 85.13.4. As $K = \text{hocolim} Y_ n[n]$ the map $K \to K'$ induces a compatible family of morphisms $Y_ n[n] \to K'$. By (1) and Lemma 85.12.9 we have $X_ n = g_{n!}K_ n$. Since $Y_0 = X_0$ we find that $K_0 \to K'_0$ being zero implies $Y_0 \to K'$ is zero. Suppose we've shown that the map $Y_ n[n] \to K'$ is zero for some $n \geq 0$. From the distinguished triangle
we get an exact sequence
As $X_{n + 1}[n + 1] = g_{n + 1!}K_{n + 1}[n + 1]$ the first group is equal to
which is zero by assumption (2). By induction we conclude all the maps $Y_ n[n] \to K'$ are zero. Consider the defining distinguished triangle
for the homotopy colimit. Arguing as above, we find that it suffices to show that
is zero for all $n \geq 0$. To see this, arguing as above, it suffices to show that
for all $n \geq 0$ which follows from condition (3). $\square$
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