Lemma 48.29.3. Let f : X \to Y be a morphism of schemes. Let r \geq 0. Assume
Y is Cohen-Macaulay (Properties, Definition 28.8.1),
f factors as X \to P \to Y where the first morphism is an immersion and the second is smooth and proper,
if x \in X and \dim (\mathcal{O}_{X, x}) \leq 1, then f is Koszul at x (More on Morphisms, Definition 37.62.2), and
if \xi is a generic point of an irreducible component of X, then we have \text{trdeg}_{\kappa (f(\xi ))} \kappa (\xi ) = r.
Then with \omega _{X/Y} = H^{-r}(f^!\mathcal{O}_ Y) there is a map
\wedge ^ r\Omega _{X/Y} \longrightarrow \omega _{X/Y}
which is an isomorphism on the locus where f is smooth.
Proof.
Let U \subset X be the open subscheme over which f is a local complete intersection morphism. Since f has relative dimension r at all generic points by assumption (4) we see that the locally constant function of Lemma 48.29.2 is constant with value r and we obtain a map
\wedge ^ r\Omega _{X/Y}|_ U = \wedge ^ r \Omega _{U/Y} \longrightarrow \omega _{U/Y} = \omega _{X/Y}|_ U
which is an isomorphism in the smooth points of f (this locus is contained in U because a smooth morphism is a local complete intersection morphism). By Lemma 48.21.5 and the assumption that Y is Cohen-Macaulay the module \omega _{X/Y} is (S_2). Since U contains all the points of codimension 1 by condition (3) and using Divisors, Lemma 31.5.11 we see that j_*\omega _{U/Y} = \omega _{X/Y}. Hence the map over U extends to X and the proof is complete.
\square
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