The Stacks project

Lemma 85.13.7. In Situation 85.3.3. Let $(K_ n, K_\varphi )$ be a simplicial system of the derived category. Assume

  1. $(K_ n, K_\varphi )$ is cartesian,

  2. $\mathop{\mathrm{Hom}}\nolimits (K_ i[t], K_ i) = 0$ for $i \geq 0$ and $t > 0$.

Then there exists a cartesian object $K$ of $D(\mathcal{C}_{total})$ whose associated simplicial system is isomorphic to $(K_ n, K_\varphi )$.

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

\[ \mathop{\mathrm{Hom}}\nolimits (X_ n, X_{n - 1}) = \mathop{\mathrm{Hom}}\nolimits (K_ n, g_ n^{-1}g_{n - 1!}K_{n - 1}) = \bigoplus \nolimits _{\varphi : [n - 1] \to [n]} \mathop{\mathrm{Hom}}\nolimits (K_ n, f_\varphi ^{-1}K_{n - 1}) \]

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

\[ \ldots \to X_2 \to X_1 \to X_0 \]

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

\[ \mathop{\mathrm{Hom}}\nolimits (X_ i[i - j - 2], X_ j) = 0\text{ for }i > j + 2 \]

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

\[ \mathop{\mathrm{Hom}}\nolimits (K_ i[i - j - 2], g_ i^{-1}g_{j!}K_ j) \]

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

\[ K = \text{hocolim} Y_ n[n] \]

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

\[ g_ m^{-1} K = \text{hocolim} g_ m^{-1}Y_ n[n] \]

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

\[ g_ m^{-1}X_ n = \bigoplus \nolimits _{\varphi : [n] \to [m]} f_\varphi ^{-1}K_ n \xrightarrow {\bigoplus K_\varphi } \bigoplus \nolimits _{\varphi : [n] \to [m]} K_ m \]

These maps define an isomorphism of complexes

\[ \xymatrix{ \ldots \ar[r] & g_ m^{-1}X_2 \ar[r] \ar[d] & g_ m^{-1}X_1 \ar[r] \ar[d] & g_ m^{-1}X_0 \ar[d] \\ \ldots \ar[r] & \bigoplus \nolimits _{\varphi : [2] \to [m]} K_ m \ar[r] & \bigoplus \nolimits _{\varphi : [1] \to [m]} K_ m \ar[r] & \bigoplus \nolimits _{\varphi : [0] \to [m]} K_ m } \]

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

\[ Y'_{m, n} = \left(\bigoplus \nolimits _{\varphi : [n] \to [m]} \mathbf{Z} \to \ldots \to \bigoplus \nolimits _{\varphi : [0] \to [m]} \mathbf{Z}\right)[-n] \otimes ^\mathbf {L}_\mathbf {Z} K_ m \]

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

\[ \mathop{\mathrm{Hom}}\nolimits (g_ m^{-1}X_ i[i - j - 1], \bigoplus \nolimits _{\varphi : [j] \to [m]} K_ m) = 0\text{ for }i > j + 1 \]

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

\[ g_ m^{-1} K = \text{hocolim} g_ m^{-1}Y_ n[n] \xrightarrow {\gamma _ m} \text{hocolim} Y'_{m, n} = K_ m \]

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

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

\[ \xymatrix{ f_\varphi ^{-1}g_ m^{-1}K \ar[d]_{f_\varphi ^{-1}\gamma _ m} \ar[r]_{K(\varphi )} & g_ n^{-1}K \ar[d]^{\gamma _ n} \\ f_\varphi ^{-1}K_ m \ar[r]^{K_\varphi } & K_ n } \]

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

\[ \xymatrix{ f_\varphi ^{-1}(\bigoplus _{\psi : [0] \to [m]} f_\psi ^{-1}K_0) \ar@{=}[r] \ar[d]_{f_\varphi ^{-1}(\bigoplus K_\psi )} & f_\varphi ^{-1}g_ m^{-1}X_0 \ar[d] \ar[r]_{X_0(\varphi )} & g_ n^{-1}X_0 \ar[d] & \bigoplus _{\chi : [0] \to [n]} f_\chi ^{-1}K_0 \ar@{=}[l] \ar[d]^{\bigoplus K_\chi } \\ f_\varphi ^{-1}(\bigoplus _{\psi : [0] \to [m]} K_ m) \ar@{=}[d] & f_\varphi ^{-1}g_ m^{-1}K \ar[d]_{f_\varphi ^{-1}\gamma _ m} \ar[r]_{K(\varphi )} & g_ n^{-1}K \ar[d]^{\gamma _ n} & \bigoplus _{\chi : [0] \to [n]} K_ n \ar@{=}[d] \\ f_\varphi ^{-1}Y'_{0, m} \ar[r] & f_\varphi ^{-1}K_ m \ar[r]^{K_\varphi } & K_ n & Y'_{0, n} \ar[l] } \]

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. $\square$

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