Represent $K_0$ by any complex $\mathcal{K}_0^\bullet$ of $\mathcal{O}_0$-modules. Apply Lemma \ref{lemma-lift-complex} with $\mathcal{G}^n = 0$ for all $n$. Denote $d : \mathcal{F}^n \to \mathcal{F}^{n + 1}$ the maps produced by the lemma. Then we see that $d \circ d : \mathcal{F}^n \to \mathcal{F}^{n + 2}$ is zero modulo $\mathcal{I}$. Since $\mathcal{F}^n$ is flat, we see that $\mathcal{I}\mathcal{F}^n = \mathcal{F}^n \otimes_{\mathcal{O}} \mathcal{I} = \mathcal{F}^n_0 \otimes_{\mathcal{O}_0} \mathcal{I}$. Hence we obtain a canonical map of complexes $$ d \circ d : \mathcal{F}_0^\bullet \longrightarrow (\mathcal{F}_0^\bullet \otimes_{\mathcal{O}_0} \mathcal{I})[2] $$ Since $\mathcal{F}_0^\bullet$ is a bounded above complex of flat $\mathcal{O}_0$-modules, it is K-flat and may be used to compute derived tensor product. Moreover, the map of complexes $\mathcal{F}_0^\bullet \to \mathcal{K}_0^\bullet$ is a quasi-isomorphism by construction. Therefore the source and target of the map just constructed represent $K_0$ and $K_0 \otimes_{\mathcal{O}_0}^\mathbf{L} \mathcal{I}[2]$ and we obtain our map $\omega(K_0)$. \medskip\noindent Let us show that this procedure is compatible with maps of complexes. Namely, let $\mathcal{L}_0^\bullet$ represent another object of $D^-(\mathcal{O}_0)$ and suppose that $$ \mathcal{K}_0^\bullet \longrightarrow \mathcal{L}_0^\bullet $$ is a map of complexes. Apply Lemma \ref{lemma-lift-complex} for the complex $\mathcal{L}_0^\bullet$, the flat modules $\mathcal{F}^n$, the maps $\mathcal{F}^n \to \mathcal{F}^{n + 1}$, and the compositions $\mathcal{F}^n \to \mathcal{K}_0^n \to \mathcal{L}_0^n$ (we apologize for the reversal of letters used). We obtain flat modules $\mathcal{G}^n$, maps $\mathcal{F}^n \to \mathcal{G}^n$, maps $\mathcal{G}^n \to \mathcal{G}^{n + 1}$, and maps $\mathcal{G}^n \to \mathcal{L}_0^n$ with all properties as in the lemma. Then it is clear that $$ \xymatrix{ \mathcal{F}_0^\bullet \ar[d] \ar[r] & (\mathcal{F}_0^\bullet \otimes_{\mathcal{O}_0} \mathcal{I})[2] \ar[d] \\ \mathcal{G}_0^\bullet \ar[r] & (\mathcal{G}_0^\bullet \otimes_{\mathcal{O}_0} \mathcal{I})[2] } $$ is a commutative diagram of complexes. \medskip\noindent To see that $\omega(K_0)$ is well defined, suppose that we have two complexes $\mathcal{K}_0^\bullet$ and $(\mathcal{K}'_0)^\bullet$ of $\mathcal{O}_0$-modules representing $K_0$ and two systems $(\mathcal{F}^n, d : \mathcal{F}^n \to \mathcal{F}^{n + 1}, \mathcal{F}^n \to \mathcal{K}_0^n)$ and $((\mathcal{F}')^n, d : (\mathcal{F}')^n \to (\mathcal{F}')^{n + 1}, (\mathcal{F}')^n \to \mathcal{K}_0^n)$ as above. Then we can choose a complex $(\mathcal{K}''_0)^\bullet$ and quasi-isomorphisms $\mathcal{K}_0^\bullet  \to (\mathcal{K}''_0)^\bullet$ and $(\mathcal{K}'_0)^\bullet  \to (\mathcal{K}''_0)^\bullet$ realizing the fact that both complexes represent $K_0$ in the derived category. Next, we apply the result of the previous paragraph to $$ (\mathcal{K}_0)^\bullet \oplus (\mathcal{K}'_0)^\bullet \longrightarrow (\mathcal{K}''_0)^\bullet $$ This produces a commutative diagram $$ \xymatrix{ \mathcal{F}_0^\bullet \oplus (\mathcal{F}'_0)^\bullet \ar[d] \ar[r] & (\mathcal{F}_0^\bullet \otimes_{\mathcal{O}_0} \mathcal{I})[2] \oplus ((\mathcal{F}'_0)^\bullet \otimes_{\mathcal{O}_0} \mathcal{I})[2] \ar[d] \\ \mathcal{G}_0^\bullet \ar[r] & (\mathcal{G}_0^\bullet \otimes_{\mathcal{O}_0} \mathcal{I})[2] } $$ Since the vertical arrows give quasi-isomorphisms on the summands we conclude the desired commutativity in $D(\mathcal{O}_0)$. \medskip\noindent Having established well-definedness, the statement on compatibility with maps is a consequence of the result in the second paragraph.