Lemma 42.59.3. Let $(S, \delta )$ be as in Situation 42.7.1. Let $f : X \to Y$ be a local complete intersection morphism of schemes locally of finite type over $S$. The bivariant class $f^!$ is independent of the choice of the factorization $f = g \circ i$ with $g$ smooth (provided one exists).
Proof. Given a second such factorization $f = g' \circ i'$ we can consider the smooth morphism $g'' : P \times _ Y P' \to Y$, the immersion $i'' : X \to P \times _ Y P'$ and the factorization $f = g'' \circ i''$. Thus we may assume that we have a diagram
\[ \xymatrix{ & P' \ar[d]^ p \ar[rd]^{g'} \\ X \ar[r]^ i \ar[ru]^{i'} & P \ar[r]^ g & Y } \]
where $p$ is a smooth morphism. Then $(g')^* = p^* \circ g^*$ (Lemma 42.14.3) and hence it suffices to show that $i^! = (i')^! \circ p^*$ in $A^*(X \to P)$. Consider the commutative diagram
\[ \xymatrix{ & X \times _ P P' \ar[d]^{\overline{p}} \ar[r]_ j & P' \ar[d]^ p \\ X \ar[ru]^ s \ar[r]^1 & X \ar[r]^ i & P } \]
where $s =(1, i')$. Then $s$ and $j$ are regular immersions (by Divisors, Lemma 31.22.8 and Divisors, Lemma 31.21.4) and $i' = j \circ s$. By Lemma 42.59.1 we have $(i')^! = s^! \circ j^!$. Since the square is cartesian, the bivariant class $j^!$ is the restriction (Remark 42.33.5) of $i^!$ to $P'$, see Lemma 42.54.2. Since bivariant classes commute with flat pullbacks we find $j^! \circ p^* = \overline{p}^* \circ i^!$. Thus it suffices to show that $s^! \circ \overline{p}^* = \text{id}$ which is done in Lemma 42.59.2. $\square$
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