Proof.
Via the identifications $\mathop{N\! L}\nolimits _ f = \tau _{\geq -1}L_ f$ (Lemma 92.22.4) and $H^0(L_ f) = \Omega _ f$ (Lemma 92.22.2) we have seen parts (2) and (3) in Deformation Theory, Lemmas 91.13.1 and 91.13.3.
Proof of (1). To match notation with Deformation Theory, Section 91.13 we will write $\mathop{N\! L}\nolimits _ f = \mathop{N\! L}\nolimits _{\mathcal{O}/\mathcal{O}_\mathcal {B}}$ and $L_ f = L_{\mathcal{O}/\mathcal{O}_\mathcal {B}}$ and similarly for the morphisms $t$ and $t \circ f$. By Deformation Theory, Lemma 91.13.8 there exists an element
\[ \xi ' \in \mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}( Lf^*\mathop{N\! L}\nolimits _{\mathcal{O}_\mathcal {B}/\mathcal{O}_{\mathcal{B}'}}, \mathcal{G}) = \mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}( Lf^*L_{\mathcal{O}_\mathcal {B}/\mathcal{O}_{\mathcal{B}'}}, \mathcal{G}) \]
such that a solution exists if and only if this element is in the image of the map
\[ \mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}( \mathop{N\! L}\nolimits _{\mathcal{O}/\mathcal{O}_{\mathcal{B}'}}, \mathcal{G}) = \mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}( L_{\mathcal{O}/\mathcal{O}_{\mathcal{B}'}}, \mathcal{G}) \longrightarrow \mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}( Lf^*L_{\mathcal{O}_\mathcal {B}/\mathcal{O}_{\mathcal{B}'}}, \mathcal{G}) \]
The distinguished triangle of Lemma 92.22.3 for $f$ and $t$ gives rise to a long exact sequence
\[ \ldots \to \mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}( L_{\mathcal{O}/\mathcal{O}_{\mathcal{B}'}}, \mathcal{G}) \to \mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}( Lf^*L_{\mathcal{O}_\mathcal {B}/\mathcal{O}_{\mathcal{B}'}}, \mathcal{G}) \to \mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {O}( L_{\mathcal{O}/\mathcal{O}_\mathcal {B}}, \mathcal{G}) \]
Hence taking $\xi $ the image of $\xi '$ works.
$\square$
Comments (0)
There are also: