Lemma 85.24.3 (BBD glueing lemma). In Situation 85.24.1. Assume

1. $\mathcal{C}$ has equalizers and fibre products,

2. there is a morphism of sites $f : \mathcal{C} \to \mathcal{D}$ given by a continuous functor $u : \mathcal{D} \to \mathcal{C}$ such that

1. $\mathcal{D}$ has equalizers and fibre products and $u$ commutes with them,

2. $\mathcal{B}$ is a full subcategory of $\mathcal{D}$ and $u : \mathcal{B} \to \mathcal{C}$ is the restriction of $u$,

3. every object of $\mathcal{D}$ has a covering whose members are objects of $\mathcal{B}$,

3. all negative self-exts of $E_ U$ in $D(\mathcal{O}_{u(U)})$ are zero, and

4. there exists a $t \in \mathbf{Z}$ such that $H^ i(E_ U) = 0$ for $i < t$ and $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{B})$.

Then there exists a solution unique up to unique isomorphism.

Proof. By Hypercoverings, Lemma 25.12.3 there exists a hypercovering $L$ for the site $\mathcal{D}$ such that $L_ n = \{ U_{n, i}\} _{i \in I_ n}$ with $U_{i, n} \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{B})$. Set $K = u(L)$. Apply Lemma 85.24.2 to get a cartesian object $E$ of $D(\mathcal{O})$ on the site $(\mathcal{C}/K)_{total}$ restricting to $E_{U_{n, i}}$ on $\mathcal{C}/u(U_{n, i})$ compatibly. The assumption on $t$ implies that $E \in D^+(\mathcal{O})$. By Hypercoverings, Lemma 25.12.4 we see that $K$ is a hypercovering too. By Lemma 85.18.4 we find that $E = a^*F$ for some $F$ in $D^+(\mathcal{O}_\mathcal {C})$.

To prove that $F$ is a solution we will use the construction of $L_0$ and $L_1$ given in the proof of Hypercoverings, Lemma 25.12.3. (This is a bit inelegant but there does not seem to be a completely straightforward way around it.)

Namely, we have $I_0 = \mathop{\mathrm{Ob}}\nolimits (\mathcal{B})$ and so $L_0 = \{ U\} _{U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{B})}$. Hence the isomorphism $a^*F \to E$ restricted to the components $\mathcal{C}/u(U)$ of $\mathcal{C}/K_0$ defines isomorphisms $\rho _ U : F|_{\mathcal{C}/u(U)} \to E_ U$ for $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{B})$ by our choice of $E$.

To prove that $\rho _ U$ satisfy the requirement of compatibility with the maps $\rho _ a$ of Situation 85.24.1 we use that $I_1$ contains the set

$\Omega = \{ (U, V, W, a, b) \mid U, V, W \in \mathcal{B}, a : U \to V, b : U \to W\}$

and that for $i = (U, V, W, a, b)$ in $\Omega$ we have $U_{1, i} = U$. Moreover, the component maps $f_{\delta ^1_0, i}$ and $f_{\delta ^1_1, i}$ of the two morphisms $K_1 \to K_0$ are the morphisms

$a : U \to V \quad \text{and}\quad b : U \to V$

Hence the compatibility mentioned in Lemma 85.24.2 gives that

$\rho _ a \circ \rho _ V|_{\mathcal{C}/u(U)} = \rho _ U \quad \text{and}\quad \rho _ b \circ \rho _ W|_{\mathcal{C}/u(U)} = \rho _ U$

Taking $i = (U, V, U, a, \text{id}_ U) \in \Omega$ for example, we find that we have the desired compatibility. The uniqueness of $F$ follows from the uniqueness of $E$ in the previous lemma (small detail omitted). $\square$

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