Lemma 91.5.5. Let (f, f') be a morphism of first order thickenings as in Situation 91.3.1. Let \mathcal{F}', \mathcal{G}' be \mathcal{O}_{X'}-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}_ X-linear map. Assume that \mathcal{F}' and \mathcal{G}' are flat over S' and that (f, f') is a strict morphism of thickenings. There exists an element
o(\varphi ) \in \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G} \otimes _{\mathcal{O}_ X} f^*\mathcal{J})
whose vanishing is a necessary and sufficient condition for the existence of a lift of \varphi to an \mathcal{O}_{X'}-linear map \varphi ' : \mathcal{F}' \to \mathcal{G}'.
First proof.
This follows from Lemma 91.5.4 as we claim that under the assumptions of the lemma we have
\mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(Li^*\mathcal{F}', \mathcal{I}\mathcal{G}') = \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G} \otimes _{\mathcal{O}_ X} f^*\mathcal{J})
Namely, we have \mathcal{I}\mathcal{G}' = \mathcal{G} \otimes _{\mathcal{O}_ X} f^*\mathcal{J} by Lemma 91.5.2. On the other hand, observe that
H^{-1}(Li^*\mathcal{F}') = \text{Tor}_1^{\mathcal{O}_{X'}}(\mathcal{F}', \mathcal{O}_ X)
(local computation omitted). Using the short exact sequence
0 \to \mathcal{I} \to \mathcal{O}_{X'} \to \mathcal{O}_ X \to 0
we see that this \text{Tor}_1 is computed by the kernel of the map \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{F} \to \mathcal{I}\mathcal{F}' which is zero by the final assertion of Lemma 91.5.2. Thus \tau _{\geq -1}Li^*\mathcal{F}' = \mathcal{F}. On the other hand, we have
\mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(Li^*\mathcal{F}', \mathcal{I}\mathcal{G}') = \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(\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 91.4.2 as follows. Note that \mathcal{K} = \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{F} and \mathcal{L} = \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{G} by Lemma 91.5.2, 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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