**Proof.**
Write $\mathcal{X} = [U/R]$ as in the proof of Lemma 102.10.1. Let $\mathcal{F}$ be an object of $\textit{LQCoh}^{fbc}(\mathcal{O}_\mathcal {X})$. It is clear from the proof of Lemma 102.10.1 that $\mathcal{F}$ is in the kernel of $Q$ if and only if $\mathcal{F}|_{U_{\acute{e}tale}} = 0$. In particular, if $\mathcal{F}$ is parasitic then $\mathcal{F}$ is in the kernel. Next, let $x : V \to \mathcal{X}$ be a flat morphism, where $V$ is a scheme. Set $W = V \times _\mathcal {X} U$ and consider the diagram

\[ \xymatrix{ W \ar[d]_ p \ar[r]_ q & V \ar[d] \\ U \ar[r] & \mathcal{X} } \]

Note that the projection $p : W \to U$ is flat and the projection $q : W \to V$ is smooth and surjective. This implies that $q_{small}^*$ is a faithful functor on quasi-coherent modules. By assumption $\mathcal{F}$ has the flat base change property so that we obtain $p_{small}^*\mathcal{F}|_{U_{\acute{e}tale}} \cong q_{small}^*\mathcal{F}|_{V_{\acute{e}tale}}$. Thus if $\mathcal{F}$ is in the kernel of $Q$, then $\mathcal{F}|_{V_{\acute{e}tale}} = 0$ which completes the proof of (1).

Part (2) follows from the discussion above and the fact that the map $Q(\mathcal{F}) \to \mathcal{F}$ becomes an isomorphism after restricting to $U_{\acute{e}tale}$.

To see part (3) note that $Q$ is left exact as a right adjoint. Let $0 \to \mathcal{F} \to \mathcal{G} \to \mathcal{H} \to 0$ be a short exact sequence in $\textit{LQCoh}^{fbc}(\mathcal{O}_\mathcal {X})$. Consider the following commutative diagram

\[ \xymatrix{ 0 \ar[r] & Q(\mathcal{F}) \ar[r] \ar[d]_ a & Q(\mathcal{G}) \ar[r] \ar[d]_ b & Q(\mathcal{H}) \ar[r] \ar[d]_ c & 0 \\ 0 \ar[r] & \mathcal{F} \ar[r] & \mathcal{G} \ar[r] & \mathcal{H} \ar[r] & 0 } \]

Since the kernels and cokernels of $a$, $b$, and $c$ are parasitic by part (2) and since the bottom row is a short exact sequence, we see that the top row as a complex of $\mathcal{O}_\mathcal {X}$-modules has parasitic cohomology sheaves (details omitted; this uses that the category of parasitic modules is a Serre subcategory of the category of all modules). By left exactness of $Q$ we see that only exactness at $Q(\mathcal{H})$ is at issue. However, the cokernel $\mathcal{Q}$ of $Q(\mathcal{G}) \to Q(\mathcal{H}))$ may be computed either in $\textit{Mod}(\mathcal{O}_\mathcal {X})$ or in $\mathit{QCoh}(\mathcal{O}_\mathcal {X})$ with the same result because the inclusion functor $\mathit{QCoh}(\mathcal{O}_\mathcal {X}) \to \textit{LQCoh}^{fbc}(\mathcal{O}_\mathcal {X})$ is a left adjoint and hence right exact. Hence $\mathcal{Q} = Q(\mathcal{Q})$ is both quasi-coherent and parasitic, whence $0$ by part (1) as desired.

As a right adjoint $Q$ commutes with all limits. Since $Q$ is exact, to show that $Q$ commutes with all colimits it suffices to show that $Q$ commutes with direct sums, see Categories, Lemma 4.14.12. Let $\mathcal{F}_ i$, $i \in I$ be a family of objects of $\textit{LQCoh}^{fbc}(\mathcal{O}_\mathcal {X})$. To see that $Q(\bigoplus \mathcal{F}_ i)$ is equal to $\bigoplus Q(\mathcal{F}_ i)$ we look at the construction of $Q$ in the proof of Lemma 102.10.1. This uses a presentation $\mathcal{X} = [U/R]$ where $U$ is a scheme. Then $Q(\mathcal{F})$ is computed by first taking the pair $(\mathcal{F}|_{U_{\acute{e}tale}}, \alpha )$ in $\mathit{QCoh}(U, R, s, t, c)$ and then using the equivalence $\mathit{QCoh}(U, R, s, t, c) \cong \mathit{QCoh}(\mathcal{O}_\mathcal {X})$. Since the restriction functor $\textit{Mod}(\mathcal{O}_\mathcal {X}) \to \textit{Mod}(\mathcal{O}_{U_{\acute{e}tale}})$, $\mathcal{F} \mapsto \mathcal{F}|_{U_{\acute{e}tale}}$ commutes with direct sums, the desired equality is clear.
$\square$

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