Lemma 91.4.2. Let i : (X, \mathcal{O}_ X) \to (X', \mathcal{O}_{X'}) be a first order thickening of ringed spaces. Assume given extensions
0 \to \mathcal{K} \to \mathcal{F}' \to \mathcal{F} \to 0 \quad \text{and}\quad 0 \to \mathcal{L} \to \mathcal{G}' \to \mathcal{G} \to 0
as in (91.4.0.1) and maps \varphi : \mathcal{F} \to \mathcal{G} and \psi : \mathcal{K} \to \mathcal{L}. Assume the diagram
\xymatrix{ \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{F} \ar[r]_-{c_{\mathcal{F}'}} \ar[d]_{1 \otimes \varphi } & \mathcal{K} \ar[d]^\psi \\ \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{G} \ar[r]^-{c_{\mathcal{G}'}} & \mathcal{L} }
is commutative. Then there exists an element
o(\varphi , \psi ) \in \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(\mathcal{F}, \mathcal{L})
whose vanishing is a necessary and sufficient condition for the existence of a map \varphi ' : \mathcal{F}' \to \mathcal{G}' compatible with \varphi and \psi .
Proof.
We can construct explicitly an extension
0 \to \mathcal{L} \to \mathcal{H} \to \mathcal{F} \to 0
by taking \mathcal{H} to be the cohomology of the complex
\mathcal{K} \xrightarrow {1, - \psi } \mathcal{F}' \oplus \mathcal{G}' \xrightarrow {\varphi , 1} \mathcal{G}
in the middle (with obvious notation). A calculation with local sections using the assumption that the diagram of the lemma commutes shows that \mathcal{H} is annihilated by \mathcal{I}. Hence \mathcal{H} defines a class in
\mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(\mathcal{F}, \mathcal{L}) \subset \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_{X'}}(\mathcal{F}, \mathcal{L})
Finally, the class of \mathcal{H} is the difference of the pushout of the extension \mathcal{F}' via \psi and the pullback of the extension \mathcal{G}' via \varphi (calculations omitted). Thus the vanishing of the class of \mathcal{H} is equivalent to the existence of a commutative diagram
\xymatrix{ 0 \ar[r] & \mathcal{K} \ar[r] \ar[d]_{\psi } & \mathcal{F}' \ar[r] \ar[d]_{\varphi '} & \mathcal{F} \ar[r] \ar[d]_\varphi & 0\\ 0 \ar[r] & \mathcal{L} \ar[r] & \mathcal{G}' \ar[r] & \mathcal{G} \ar[r] & 0 }
as desired.
\square
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