**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).
$\square$

## Comments (0)

There are also: