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
Comments (0)