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. \medskip\noindent 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 \ref{lemma-induced-map}. 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 \ref{homology-lemma-four-lemma} we see that $H^{n - 1}(\delta)$ is surjective. This in turn implies that $H^{n - 1}(\beta)$ is surjective by Lemma \ref{lemma-induced-map}. Using Homology, Lemma \ref{homology-lemma-four-lemma} 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.