Lemma 91.11.6. Let $(f, f')$ be a morphism of first order thickenings as in Situation 91.9.1. Let $\mathcal{F}'$, $\mathcal{G}'$ be $\mathcal{O}'$-modules and set $\mathcal{F} = i^*\mathcal{F}'$ and $\mathcal{G} = i^*\mathcal{G}'$. Let $\varphi : \mathcal{F} \to \mathcal{G}$ be an $\mathcal{O}$-linear map. Assume that $\mathcal{F}'$ and $\mathcal{G}'$ are flat over $\mathcal{O}_{\mathcal{B}'}$ and that $(f, f')$ is a strict morphism of thickenings. There exists an element
\[ o(\varphi ) \in \mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}(\mathcal{F}, \mathcal{G} \otimes _\mathcal {O} f^*\mathcal{J}) \]
whose vanishing is a necessary and sufficient condition for the existence of a lift of $\varphi $ to an $\mathcal{O}'$-linear map $\varphi ' : \mathcal{F}' \to \mathcal{G}'$.
First proof.
This follows from Lemma 91.11.5 as we claim that under the assumptions of the lemma we have
\[ \mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}(Li^*\mathcal{F}', \mathcal{I}\mathcal{G}') = \mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}(\mathcal{F}, \mathcal{G} \otimes _\mathcal {O} f^*\mathcal{J}) \]
Namely, we have $\mathcal{I}\mathcal{G}' = \mathcal{G} \otimes _\mathcal {O} f^*\mathcal{J}$ by Lemma 91.11.2. On the other hand, observe that
\[ H^{-1}(Li^*\mathcal{F}') = \text{Tor}_1^{\mathcal{O}'}(\mathcal{F}', \mathcal{O}) \]
(local computation omitted). Using the short exact sequence
\[ 0 \to \mathcal{I} \to \mathcal{O}' \to \mathcal{O} \to 0 \]
we see that this $\text{Tor}_1$ is computed by the kernel of the map $\mathcal{I} \otimes _\mathcal {O} \mathcal{F} \to \mathcal{I}\mathcal{F}'$ which is zero by the final assertion of Lemma 91.11.2. Thus $\tau _{\geq -1}Li^*\mathcal{F}' = \mathcal{F}$. On the other hand, we have
\[ \mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}(Li^*\mathcal{F}', \mathcal{I}\mathcal{G}') = \mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}(\tau _{\geq -1}Li^*\mathcal{F}', \mathcal{I}\mathcal{G}') \]
by the dual of Derived Categories, Lemma 13.16.1.
$\square$
Second proof.
We can apply Lemma 91.10.2 as follows. Note that $\mathcal{K} = \mathcal{I} \otimes _\mathcal {O} \mathcal{F}$ and $\mathcal{L} = \mathcal{I} \otimes _\mathcal {O} \mathcal{G}$ by Lemma 91.11.2, that $c_{\mathcal{F}'} = 1 \otimes 1$ and $c_{\mathcal{G}'} = 1 \otimes 1$ and taking $\psi = 1 \otimes \varphi $ the diagram of the lemma commutes. Thus $o(\varphi ) = o(\varphi , 1 \otimes \varphi )$ works.
$\square$
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