Proof.
An automorphism of $(K, \alpha )$ is a map $\varphi : K \to K$ in $D(\mathcal{O})$ with $\varphi \otimes _\mathcal {O} \text{id}_{\mathcal{O}_0} = \text{id}$. This is the same thing as saying that
\[ K \xrightarrow {\varphi - \text{id}} K \to K \otimes _\mathcal {O}^\mathbf {L} \mathcal{O}_0 \]
is zero. We conclude the group of automorphisms is the cokernel of a map
\[ \mathop{\mathrm{Hom}}\nolimits _\mathcal {O}(K, K_0[-1]) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _\mathcal {O}(K, K_0 \otimes _{\mathcal{O}_0}^\mathbf {L} \mathcal{I}) \]
by the distinguished triangle
\[ K \otimes _\mathcal {O}^\mathbf {L} \mathcal{I} \to K \to K \otimes _\mathcal {O}^\mathbf {L} \mathcal{O}_0 \to (K \otimes _\mathcal {O}^\mathbf {L} \mathcal{I})[1] \]
in $D(\mathcal{O})$ and Derived Categories, Lemma 13.4.2. To translate into the groups in the lemma use adjunction of the restriction functor $D(\mathcal{O}_0) \to D(\mathcal{O})$ and $- \otimes _\mathcal {O} \mathcal{O}_0 : D(\mathcal{O}) \to D(\mathcal{O}_0)$. This proves (1).
Proof of (2). Assume that $K_0 = K \otimes _\mathcal {O}^\mathbf {L} \mathcal{O}_0$ in $D(\mathcal{O})$. By Lemma 91.16.6 the map sending a lift $(K', \alpha _0)$ to the obstruction $o(\alpha _0)$ to lifting $\alpha _0$ defines a canonical injective map from the set of isomomorphism classes of pairs to $\mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_0}(K_0, K_0 \otimes _{\mathcal{O}_0}^\mathbf {L} \mathcal{I})$. To finish the proof we show that it is surjective. Pick $\xi : K_0 \to (K_0 \otimes _{\mathcal{O}_0}^\mathbf {L} \mathcal{I})[1]$ in the $\mathop{\mathrm{Ext}}\nolimits ^1$ of the lemma. Choose a bounded above complex $\mathcal{F}^\bullet $ of flat $\mathcal{O}$-modules representing $K$. The map $\xi $ can be represented as $t \circ s^{-1}$ where $s : \mathcal{K}_0^\bullet \to \mathcal{F}_0^\bullet $ is a quasi-isomorphism and $t : \mathcal{K}_0^\bullet \to \mathcal{F}_0^\bullet \otimes _{\mathcal{O}_0} \mathcal{I}[1]$ is a map of complexes. By Lemma 91.16.5 we can assume there exists a quasi-isomorphism $\mathcal{G}^\bullet \to \mathcal{F}^\bullet $ of complexes of $\mathcal{O}$-modules such that $\mathcal{G}_0^\bullet \to \mathcal{F}_0^\bullet $ factors through $s$ up to homotopy. We may and do replace $\mathcal{G}^\bullet $ by a bounded above complex of flat $\mathcal{O}$-modules (by picking a qis from such to $\mathcal{G}^\bullet $ and replacing). Then we see that $\xi $ is represented by a map of complexes $t : \mathcal{G}_0^\bullet \to \mathcal{F}_0^\bullet \otimes _{\mathcal{O}_0} \mathcal{I}[1]$ and the quasi-isomorphism $\mathcal{G}_0^\bullet \to \mathcal{F}_0^\bullet $. Set
\[ \mathcal{H}^ n = \mathcal{F}^ n \times _{\mathcal{F}_0^ n} \mathcal{G}_0^ n \]
with differentials
\[ \mathcal{H}^ n \to \mathcal{H}^{n + 1},\quad (f^ n, g_0^ n) \mapsto (d(f^ n) + t(g_0^ n), d(g_0^ n)) \]
This makes sense as $\mathcal{F}_0^{n + 1} \otimes _{\mathcal{O}_0} \mathcal{I} = \mathcal{F}^{n + 1} \otimes _\mathcal {O} \mathcal{I} = \mathcal{I}\mathcal{F}^{n + 1} \subset \mathcal{F}^{n + 1}$. We omit the computation that shows that $\mathcal{H}^\bullet $ is a complex of $\mathcal{O}$-modules. By construction there is a short exact sequence
\[ 0 \to \mathcal{F}_0^\bullet \otimes _{\mathcal{O}_0} \mathcal{I} \to \mathcal{H}^\bullet \to \mathcal{G}_0^\bullet \to 0 \]
of complexes of $\mathcal{O}$-modules. Exactly as in the proof of Lemma 91.16.4 one shows that this sequence induces an isomorphism $\alpha _0 : \mathcal{H}^\bullet \otimes _\mathcal {O}^\mathbf {L} \mathcal{O}_0 \to \mathcal{G}_0^\bullet $ in $D(\mathcal{O}_0)$. In other words, we have produced a pair $(\mathcal{H}^\bullet , \alpha _0)$. We omit the verification that $o(\alpha _0) = \xi $; hint: $o(\alpha _0)$ can be computed explitly in this case as we have maps $\mathcal{H}^ n \to \mathcal{F}^ n$ (not compatible with differentials) lifting the components of $\alpha _0$. This finishes the proof.
$\square$
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