Lemma 91.4.4. Let i : (X, \mathcal{O}_ X) \to (X', \mathcal{O}_{X'}) be a first order thickening of ringed spaces. Assume given \mathcal{O}_ X-modules \mathcal{F}, \mathcal{K} and an \mathcal{O}_ X-linear map c : \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{F} \to \mathcal{K}. Then there exists an element
o(\mathcal{F}, \mathcal{K}, c) \in \mathop{\mathrm{Ext}}\nolimits ^2_{\mathcal{O}_ X}(\mathcal{F}, \mathcal{K})
whose vanishing is a necessary and sufficient condition for the existence of a sequence (91.4.0.1) with c_{\mathcal{F}'} = c.
Proof.
We first show that if \mathcal{K} is an injective \mathcal{O}_ X-module, then there does exist a sequence (91.4.0.1) with c_{\mathcal{F}'} = c. To do this, choose a flat \mathcal{O}_{X'}-module \mathcal{H}' and a surjection \mathcal{H}' \to \mathcal{F} (Modules, Lemma 17.17.6). Let \mathcal{J} \subset \mathcal{H}' be the kernel. Since \mathcal{H}' is flat we have
\mathcal{I} \otimes _{\mathcal{O}_{X'}} \mathcal{H}' = \mathcal{I}\mathcal{H}' \subset \mathcal{J} \subset \mathcal{H}'
Observe that the map
\mathcal{I}\mathcal{H}' = \mathcal{I} \otimes _{\mathcal{O}_{X'}} \mathcal{H}' \longrightarrow \mathcal{I} \otimes _{\mathcal{O}_{X'}} \mathcal{F} = \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{F}
annihilates \mathcal{I}\mathcal{J}. Namely, if f is a local section of \mathcal{I} and s is a local section of \mathcal{H}, then fs is mapped to f \otimes \overline{s} where \overline{s} is the image of s in \mathcal{F}. Thus we obtain
\xymatrix{ \mathcal{I}\mathcal{H}'/\mathcal{I}\mathcal{J} \ar@{^{(}->}[r] \ar[d] & \mathcal{J}/\mathcal{I}\mathcal{J} \ar@{..>}[d]_\gamma \\ \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{F} \ar[r]^-c & \mathcal{K} }
a diagram of \mathcal{O}_ X-modules. If \mathcal{K} is injective as an \mathcal{O}_ X-module, then we obtain the dotted arrow. Denote \gamma ' : \mathcal{J} \to \mathcal{K} the composition of \gamma with \mathcal{J} \to \mathcal{J}/\mathcal{I}\mathcal{J}. A local calculation shows the pushout
\xymatrix{ 0 \ar[r] & \mathcal{J} \ar[r] \ar[d]_{\gamma '} & \mathcal{H}' \ar[r] \ar[d] & \mathcal{F} \ar[r] \ar@{=}[d] & 0 \\ 0 \ar[r] & \mathcal{K} \ar[r] & \mathcal{F}' \ar[r] & \mathcal{F} \ar[r] & 0 }
is a solution to the problem posed by the lemma.
General case. Choose an embedding \mathcal{K} \subset \mathcal{K}' with \mathcal{K}' an injective \mathcal{O}_ X-module. Let \mathcal{Q} be the quotient, so that we have an exact sequence
0 \to \mathcal{K} \to \mathcal{K}' \to \mathcal{Q} \to 0
Denote c' : \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{F} \to \mathcal{K}' be the composition. By the paragraph above there exists a sequence
0 \to \mathcal{K}' \to \mathcal{E}' \to \mathcal{F} \to 0
as in (91.4.0.1) with c_{\mathcal{E}'} = c'. Note that c' composed with the map \mathcal{K}' \to \mathcal{Q} is zero, hence the pushout of \mathcal{E}' by \mathcal{K}' \to \mathcal{Q} is an extension
0 \to \mathcal{Q} \to \mathcal{D}' \to \mathcal{F} \to 0
as in (91.4.0.1) with c_{\mathcal{D}'} = 0. This means exactly that \mathcal{D}' is annihilated by \mathcal{I}, in other words, the \mathcal{D}' is an extension of \mathcal{O}_ X-modules, i.e., defines an element
o(\mathcal{F}, \mathcal{K}, c) \in \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(\mathcal{F}, \mathcal{Q}) = \mathop{\mathrm{Ext}}\nolimits ^2_{\mathcal{O}_ X}(\mathcal{F}, \mathcal{K})
(the equality holds by the long exact cohomology sequence associated to the exact sequence above and the vanishing of higher ext groups into the injective module \mathcal{K}'). If o(\mathcal{F}, \mathcal{K}, c) = 0, then we can choose a splitting s : \mathcal{F} \to \mathcal{D}' and we can set
\mathcal{F}' = \mathop{\mathrm{Ker}}(\mathcal{E}' \to \mathcal{D}'/s(\mathcal{F}))
so that we obtain the following diagram
\xymatrix{ 0 \ar[r] & \mathcal{K} \ar[r] \ar[d] & \mathcal{F}' \ar[r] \ar[d] & \mathcal{F} \ar[r] \ar@{=}[d] & 0 \\ 0 \ar[r] & \mathcal{K}' \ar[r] & \mathcal{E}' \ar[r] & \mathcal{F} \ar[r] & 0 }
with exact rows which shows that c_{\mathcal{F}'} = c. Conversely, if \mathcal{F}' exists, then the pushout of \mathcal{F}' by the map \mathcal{K} \to \mathcal{K}' is isomorphic to \mathcal{E}' by Lemma 91.4.3 and the vanishing of higher ext groups into the injective module \mathcal{K}'. This gives a diagram as above, which implies that \mathcal{D}' is split as an extension, i.e., the class o(\mathcal{F}, \mathcal{K}, c) is zero.
\square
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