Proof.
Let $K$ be as in (2). Then we can represent $K$ by a bounded above complex $\mathcal{F}^\bullet $ of flat $\mathcal{O}$-modules. Then $\mathcal{F}_0^\bullet = \mathcal{F}^\bullet \otimes _{\mathcal{O}} \mathcal{O}_0$ represents $K_0$ in $D(\mathcal{O}_0)$. Since $d_{\mathcal{F}^\bullet } \circ d_{\mathcal{F}^\bullet } = 0$ as $\mathcal{F}^\bullet $ is a complex, we see from the very construction of $\omega (K_0)$ that it is zero.
Assume (1). Let $\mathcal{F}^ n$, $d : \mathcal{F}^ n \to \mathcal{F}^{n + 1}$ be as in the construction of $\omega (K_0)$. The nullity of $\omega (K_0)$ implies that the map
\[ \omega = d \circ d : \mathcal{F}_0^\bullet \longrightarrow (\mathcal{F}_0^\bullet \otimes _{\mathcal{O}_0} \mathcal{I})[2] \]
is zero in $D(\mathcal{O}_0)$. By definition of the derived category as the localization of the homotopy category of complexes of $\mathcal{O}_0$-modules, there exists a quasi-isomorphism $\alpha : \mathcal{G}_0^\bullet \to \mathcal{F}_0^\bullet $ such that there exist $\mathcal{O}_0$-modules maps $h^ n : \mathcal{G}_0^ n \to \mathcal{F}_0^{n + 1} \otimes _\mathcal {O} \mathcal{I}$ with
\[ \omega \circ \alpha = d_{\mathcal{F}_0^\bullet \otimes \mathcal{I}} \circ h + h \circ d_{\mathcal{G}_0^\bullet } \]
We set
\[ \mathcal{H}^ n = \mathcal{F}^ n \times _{\mathcal{F}^ n_0} \mathcal{G}_0^ n \]
and we define
\[ d' : \mathcal{H}^ n \longrightarrow \mathcal{H}^{n + 1},\quad (f^ n, g_0^ n) \longmapsto (d(f^ n) - h^ n(g_0^ n), d(g_0^ n)) \]
with obvious notation using that $\mathcal{F}_0^{n + 1} \otimes _{\mathcal{O}_0} \mathcal{I} = \mathcal{F}^{n + 1} \otimes _\mathcal {O} \mathcal{I} = \mathcal{I}\mathcal{F}^{n + 1} \subset \mathcal{F}^{n + 1}$. Then one checks $d' \circ d' = 0$ by our choice of $h^ n$ and definition of $\omega $. Hence $\mathcal{H}^\bullet $ defines an object in $D(\mathcal{O})$. On the other hand, there is a short exact sequence of complexes of $\mathcal{O}$-modules
\[ 0 \to \mathcal{F}_0^\bullet \otimes _{\mathcal{O}_0} \mathcal{I} \to \mathcal{H}^\bullet \to \mathcal{G}_0^\bullet \to 0 \]
We still have to show that $\mathcal{H}^\bullet \otimes _\mathcal {O}^\mathbf {L} \mathcal{O}_0$ is isomorphic to $K_0$. Choose a quasi-isomorphism $\mathcal{E}^\bullet \to \mathcal{H}^\bullet $ where $\mathcal{E}^\bullet $ is a bounded above complex of flat $\mathcal{O}$-modules. We obtain a commutative diagram
\[ \xymatrix{ 0 \ar[r] & \mathcal{E}^\bullet \otimes _\mathcal {O} \mathcal{I} \ar[d]^\beta \ar[r] & \mathcal{E}^\bullet \ar[d]^\gamma \ar[r] & \mathcal{E}_0^\bullet \ar[d]^\delta \ar[r] & 0 \\ 0 \ar[r] & \mathcal{F}_0^\bullet \otimes _{\mathcal{O}_0} \mathcal{I} \ar[r] & \mathcal{H}^\bullet \ar[r] & \mathcal{G}_0^\bullet \ar[r] & 0 } \]
We claim that $\delta $ is a quasi-isomorphism. Since $H^ i(\delta )$ is an isomorphism for $i \gg 0$, we can use descending induction on $n$ such that $H^ i(\delta )$ is an isomorphism for $i \geq n$. Observe that $\mathcal{E}^\bullet \otimes _\mathcal {O} \mathcal{I}$ represents $\mathcal{E}_0^\bullet \otimes _{\mathcal{O}_0}^\mathbf {L} \mathcal{I}$, that $\mathcal{F}_0^\bullet \otimes _{\mathcal{O}_0} \mathcal{I}$ represents $\mathcal{G}_0^\bullet \otimes _{\mathcal{O}_0}^\mathbf {L} \mathcal{I}$, and that $\beta = \delta \otimes _{\mathcal{O}_0}^\mathbf {L} \text{id}_\mathcal {I}$ as maps in $D(\mathcal{O}_0)$. This is true because $\beta = (\alpha \otimes \text{id}_\mathcal {I}) \circ (\delta \otimes \text{id}_\mathcal {I})$. Suppose that $H^ i(\delta )$ is an isomorphism in degrees $\geq n$. Then the same is true for $\beta $ by what we just said and Lemma 91.16.3. Then we can look at the diagram
\[ \xymatrix{ H^{n - 1}(\mathcal{E}^\bullet \otimes _\mathcal {O} \mathcal{I}) \ar[r] \ar[d]^{H^{n - 1}(\beta )} & H^{n - 1}(\mathcal{E}^\bullet ) \ar[r] \ar[d] & H^{n - 1}(\mathcal{E}_0^\bullet ) \ar[r] \ar[d]^{H^{n - 1}(\delta )} & H^ n(\mathcal{E}^\bullet \otimes _\mathcal {O} \mathcal{I}) \ar[r] \ar[d]^{H^ n(\beta )} & H^ n(\mathcal{E}^\bullet ) \ar[d] \\ H^{n - 1}(\mathcal{F}_0^\bullet \otimes _\mathcal {O} \mathcal{I}) \ar[r] & H^{n - 1}(\mathcal{H}^\bullet ) \ar[r] & H^{n - 1}(\mathcal{G}_0^\bullet ) \ar[r] & H^ n(\mathcal{F}_0^\bullet \otimes _\mathcal {O} \mathcal{I}) \ar[r] & H^ n(\mathcal{H}^\bullet ) } \]
Using Homology, Lemma 12.5.19 we see that $H^{n - 1}(\delta )$ is surjective. This in turn implies that $H^{n - 1}(\beta )$ is surjective by Lemma 91.16.3. Using Homology, Lemma 12.5.19 again we see that $H^{n - 1}(\delta )$ is an isomorphism. The claim holds by induction, so $\delta $ is a quasi-isomorphism which is what we wanted to show.
$\square$
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