The Stacks project

Proof. This is obvious. $\square$

Proof. Please compare with Lemma 85.12.4 and its proof (also to see the cocycle condition spelled out). The construction is analogous to the construction discussed in Descent, Section 35.3 from which we borrow the notation $\tau ^ n_ i : [0] \to [n]$, $0 \mapsto i$ and $\tau ^ n_{ij} : [1] \to [n]$, $0 \mapsto i$, $1 \mapsto j$. Given $\varphi : [n] \to [m]$ we define $K_\varphi : f_\varphi ^{-1}K_ n \to K_ m$ using

We omit the verification that the cocycle condition implies the maps compose correctly (in their respective derived categories) and hence give rise to a simplicial system in the derived category. $\square$

Proof. First, if $K = \mathbf{Z}$, then this is the construction of Derived Categories, Example 13.41.2 applied to the complex

in $\textit{Ab}(\mathcal{C}_{total})$ combined with the fact that this complex represents $K = \mathbf{Z}$ in $D(\mathcal{C}_{total})$ by Lemma 85.8.1. The general case follows from this, the fact that the exact functor $- \otimes ^\mathbf {L}_\mathbf {Z} K$ sends Postnikov systems to Postnikov systems, and that $- \otimes ^\mathbf {L}_\mathbf {Z} K$ commutes with homotopy colimits. $\square$

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). $\square$

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. $\square$

Proof. Set $X_ n = g_{n!}K_ n$ in $D(\mathcal{C}_{total})$. For each $n \geq 1$ we have

Thus we get a map $X_ n \to X_{n - 1}$ corresponding to the alternating sum of the maps $K_\varphi ^{-1} : K_ n \to f_\varphi ^{-1}K_{n - 1}$ where $\varphi $ runs over $\delta ^ n_0, \ldots , \delta ^ n_ n$. We can do this because $K_\varphi $ is invertible by assumption (1). Please observe the similarity with the definition of the maps in the proof of Lemma 85.8.1. We obtain a complex

in $D(\mathcal{C}_{total})$. We omit the computation which shows that the compositions are zero. By Derived Categories, Lemma 13.41.6 if we have

then we can extend this complex to a Postnikov system. The group is equal to

Again using that $(K_ n, K_\varphi )$ is cartesian we see that $g_ i^{-1}g_{j!}K_ j$ is isomorphic to a finite direct sum of copies of $K_ i$. Hence the group vanishes by assumption (2). Let the Postnikov system be given by $Y_0 = X_0$ and distinguished sequences $Y_ n \to X_ n \to Y_{n - 1} \to Y_ n[1]$ for $n \geq 1$. We set

To finish the proof we have to show that $g_ m^{-1}K$ is isomorphic to $K_ m$ for all $m$ compatible with the maps $K_\varphi $. Observe that

and that $g_ m^{-1}Y_ n[n]$ is a Postnikov system for $g_ m^{-1}X_ n$. Consider the isomorphisms

These maps define an isomorphism of complexes

in $D(\mathcal{C}_ m)$ where the arrows in the bottom row are as in the proof of Lemma 85.8.1. The squares commute by our choice of the arrows of the complex $\ldots \to X_2 \to X_1 \to X_0$; we omit the computation. The bottom row complex has a postnikov tower given by

and $\text{hocolim} Y'_{m, n} = K_ m$ (please compare with the proof of Lemma 85.13.4 and Derived Categories, Example 13.41.2). Applying the second part of Derived Categories, Lemma 13.41.6 the vertical maps in the big diagram extend to an isomorphism of Postnikov systems provided we have

The is true if $\mathop{\mathrm{Hom}}\nolimits (K_ m[i - j - 1], K_ m) = 0$ for $i > j + 1$ which holds by assumption (2). Choose an isomorphism given by $\gamma _{m, n} : g_ m^{-1}Y_ n \to Y'_{m, n}$ of Postnikov systems in $D(\mathcal{C}_ m)$. By uniqueness of homotopy colimits, we can find an isomorphism

compatible with $\gamma _{m, n}$.

We still have to prove that the maps $\gamma _ m$ fit into commutative diagrams

for every $\varphi : [m] \to [n]$. Consider the diagram

The top middle square is commutative as $X_0 \to K$ is a morphism of simplicial objects. The left, resp. the right rectangles are commutative as $\gamma _ m$, resp. $\gamma _ n$ is compatible with $\gamma _{0, m}$, resp. $\gamma _{0, n}$ which are the arrows $\bigoplus K_\psi $ and $\bigoplus K_\chi $ in the diagram. Going around the outer rectangle of the diagram is commutative as $(K_ n, K_\varphi )$ is a simplical system and the map $X_0(\varphi )$ is given by the obvious identifications $f_\varphi ^{-1}f_\psi ^{-1}K_0 = f_{\varphi \circ \psi }^{-1}K_0$. Note that the arrow $\bigoplus _\psi K_ m \to Y'_{0, m} \to K_ m$ induces an isomorphism on any of the direct summands (because of our explicit construction of the Postnikov systems $Y'_{i, j}$ above). Hence, if we take a direct summand of the upper left and corner, then this maps isomorphically to $f_\varphi ^{-1}g_ m^{-1}K$ as $\gamma _ m$ is an isomorphism. Working out what the above says, but looking only at this direct summand we conclude the lower middle square commutes as we well. This concludes the proof. $\square$