Proof.
By composition we mean the map
\[ R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(\mathcal{O}_ X, \mathcal{O}_ X) \otimes _{\mathcal{O}_ Y}^\mathbf {L} R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(\mathcal{O}_ X, \mathcal{O}_ X) \longrightarrow R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(\mathcal{O}_ X, \mathcal{O}_ X) \]
of Cohomology, Lemma 20.42.5. This induces multiplication maps
\[ \mathop{\mathcal{E}\! \mathit{xt}}\nolimits ^ a_{\mathcal{O}_ Y}(\mathcal{O}_ X, \mathcal{O}_ X) \otimes _{\mathcal{O}_ Y} \mathop{\mathcal{E}\! \mathit{xt}}\nolimits ^ b_{\mathcal{O}_ Y}(\mathcal{O}_ X, \mathcal{O}_ X) \longrightarrow \mathop{\mathcal{E}\! \mathit{xt}}\nolimits ^{a + b}_{\mathcal{O}_ Y}(\mathcal{O}_ X, \mathcal{O}_ X) \]
Please compare with More on Algebra, Equation (15.63.0.1). The statement of the lemma means that the induced map
\[ \mathop{\mathcal{E}\! \mathit{xt}}\nolimits ^1_{\mathcal{O}_ Y}(\mathcal{O}_ X, \mathcal{O}_ X) \otimes \ldots \otimes \mathop{\mathcal{E}\! \mathit{xt}}\nolimits ^1_{\mathcal{O}_ Y}(\mathcal{O}_ X, \mathcal{O}_ X) \longrightarrow \mathop{\mathcal{E}\! \mathit{xt}}\nolimits ^ i_{\mathcal{O}_ Y}(\mathcal{O}_ X, \mathcal{O}_ X) \]
factors through the wedge product and then induces an isomorphism. To see this is true we may work locally on $Y$. Hence we may assume that we have global sections $f_1, \ldots , f_ r$ of $\mathcal{O}_ Y$ which generate $\mathcal{I}$ and which form a Koszul regular sequence. Denote
\[ \mathcal{A} = \mathcal{O}_ Y\langle \xi _1, \ldots , \xi _ r\rangle \]
the sheaf of strictly commutative differential graded $\mathcal{O}_ Y$-algebras which is a (divided power) polynomial algebra on $\xi _1, \ldots , \xi _ r$ in degree $-1$ over $\mathcal{O}_ Y$ with differential $\text{d}$ given by the rule $\text{d}\xi _ i = f_ i$. Let us denote $\mathcal{A}^\bullet $ the underlying complex of $\mathcal{O}_ Y$-modules which is the Koszul complex mentioned above. Thus the canonical map $\mathcal{A}^\bullet \to \mathcal{O}_ X$ is a quasi-isomorphism. We obtain quasi-isomorphisms
\[ R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(\mathcal{O}_ X, \mathcal{O}_ X) \to \mathop{\mathcal{H}\! \mathit{om}}\nolimits ^\bullet (\mathcal{A}^\bullet , \mathcal{A}^\bullet ) \to \mathop{\mathcal{H}\! \mathit{om}}\nolimits ^\bullet (\mathcal{A}^\bullet , \mathcal{O}_ X) \]
by Cohomology, Lemma 20.46.9. The differentials of the latter complex are zero, and hence
\[ \mathop{\mathcal{E}\! \mathit{xt}}\nolimits ^ i_{\mathcal{O}_ Y}(\mathcal{O}_ X, \mathcal{O}_ X) \cong \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(\mathcal{A}^{-i}, \mathcal{O}_ X) \]
For $j \in \{ 1, \ldots , r\} $ let $\delta _ j : \mathcal{A} \to \mathcal{A}$ be the derivation of degree $1$ with $\delta _ j(\xi _ i) = \delta _{ij}$ (Kronecker delta). A computation shows that $\delta _ j \circ \text{d} = - \text{d} \circ \delta _ j$ which shows that we get a morphism of complexes
\[ \delta _ j : \mathcal{A}^\bullet \to \mathcal{A}^\bullet [1]. \]
Whence $\delta _ j$ defines a section of the corresponding $\mathop{\mathcal{E}\! \mathit{xt}}\nolimits $-sheaf. Another computation shows that $\delta _1, \ldots , \delta _ r$ map to a basis for $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(\mathcal{A}^{-1}, \mathcal{O}_ X)$ over $\mathcal{O}_ X$. Since it is clear that $\delta _ j \circ \delta _ j = 0$ and $\delta _ j \circ \delta _{j'} = - \delta _{j'} \circ \delta _ j$ as endomorphisms of $\mathcal{A}$ and hence in the $\mathop{\mathcal{E}\! \mathit{xt}}\nolimits $-sheaves we obtain the statement that our map above factors through the exterior power. To see we get the desired isomorphism the reader checks that the elements
\[ \delta _{j_1} \circ \ldots \circ \delta _{j_ i} \]
for $j_1 < \ldots < j_ i$ map to a basis of the sheaf $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(\mathcal{A}^{-i}, \mathcal{O}_ X)$ over $\mathcal{O}_ X$.
$\square$
Comments (4)
Comment #2105 by Pieter Belmans on
Comment #2131 by Johan on
Comment #8643 by nkym on
Comment #9408 by Stacks project on