Lemma 17.30.5. Let $X$ be a topological space. Let $\mathcal{A} \to \mathcal{B} \to \mathcal{C}$ be maps of sheaves of rings. Let $C$ be the cone (Derived Categories, Definition 13.9.1) of the map of complexes $\mathop{N\! L}\nolimits _{\mathcal{C}/\mathcal{A}} \to \mathop{N\! L}\nolimits _{\mathcal{C}/\mathcal{B}}$. There is a canonical map

$c : \mathop{N\! L}\nolimits _{\mathcal{B}/\mathcal{A}} \otimes _\mathcal {B} \mathcal{C} \longrightarrow C[-1]$

of complexes of $\mathcal{C}$-modules which produces a canonical six term exact sequence

$\xymatrix{ H^0(\mathop{N\! L}\nolimits _{\mathcal{B}/\mathcal{A}} \otimes _\mathcal {B} \mathcal{C}) \ar[r] & H^0(\mathop{N\! L}\nolimits _{\mathcal{C}/\mathcal{A}}) \ar[r] & H^0(\mathop{N\! L}\nolimits _{\mathcal{C}/\mathcal{B}}) \ar[r] & 0 \\ H^{-1}(\mathop{N\! L}\nolimits _{\mathcal{B}/\mathcal{A}} \otimes _\mathcal {B} \mathcal{C}) \ar[r] & H^{-1}(\mathop{N\! L}\nolimits _{\mathcal{C}/\mathcal{A}}) \ar[r] & H^{-1}(\mathop{N\! L}\nolimits _{\mathcal{C}/\mathcal{B}}) \ar[llu] }$

of cohomology sheaves.

Proof. To give the map $c$ we have to give a map $c_1 : \mathop{N\! L}\nolimits _{\mathcal{B}/\mathcal{A}} \otimes _\mathcal {B} \mathcal{C} \to \mathop{N\! L}\nolimits _{\mathcal{C}/\mathcal{A}}$ and an explicit homotopy between the composition

$\mathop{N\! L}\nolimits _{\mathcal{B}/\mathcal{A}} \otimes _\mathcal {B} \mathcal{C} \to \mathop{N\! L}\nolimits _{\mathcal{C}/\mathcal{A}} \to \mathop{N\! L}\nolimits _{\mathcal{C}/\mathcal{B}}$

and the zero map, see Derived Categories, Lemma 13.9.3. For $c_1$ we use the functoriality described above for the obvious diagram. For the homotopy we use the map

$\mathop{N\! L}\nolimits _{\mathcal{B}/\mathcal{A}}^0 \otimes _\mathcal {B} \mathcal{C} \longrightarrow \mathop{N\! L}\nolimits _{\mathcal{C}/\mathcal{B}}^{-1},\quad \text{d}[b] \otimes 1 \longmapsto [\varphi (b)] - b[1]$

where $\varphi : \mathcal{B} \to \mathcal{C}$ is the given map. Please compare with Algebra, Remark 10.134.5. To see the consequence for cohomology sheaves, it suffices to show that $H^0(c)$ is an isomorphism and $H^{-1}(c)$ surjective. To see this we can look at stalks, see Lemma 17.30.4, and then we can use the corresponding result in commutative algebra, see Algebra, Lemma 10.134.4. Some details omitted. $\square$

Comment #7433 by nkym on

I found a small typo "explicity" in the proof.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).