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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