The Stacks project

Proof. The fact that $\mathcal{C}$ is finite locally free follows immediately from More on Algebra, Lemma 15.85.5. Choose a factorization $f = h \circ i$ with $h : P \to Y$ smooth and $i$ an immersion. Then we can factor $f' = h' \circ i'$ where $h' : P' \to Y'$ and $i' : X' \to P'$ the base changes. Picture

In particular, we see that the gysin map exists for $f'$. By More on Morphisms, Lemmas 37.13.13 we have

where $\mathcal{C}_{X/P}$ is the conormal sheaf of the embedding $i$. Similarly for the primed version. We have $(g')^*i^*\Omega _{P/Y} = (i')^*\Omega _{P'/Y'}$ because $\Omega _{P/Y}$ pulls back to $\Omega _{P'/Y'}$ by Morphisms, Lemma 29.32.10. Also, recall that $(g')^*\mathcal{C}_{X/P} \to \mathcal{C}_{X'/P'}$ is surjective, see Morphisms, Lemma 29.31.4. We deduce that the sheaf $\mathcal{C}$ is canonicallly isomorphic to the kernel of the map $(g')^*\mathcal{C}_{X/P} \to \mathcal{C}_{X'/P'}$ of finite locally free modules. Recall that $i^!$ is defined using $\mathcal{N} = \mathcal{C}_{Z/X}^\vee$ and similarly for $(i')^!$. Thus we have

in $A^*(X' \to P')$ by an application of Lemma 42.54.4. Since finally we have $f^! = i^! \circ h^*$, $(f')^! = (i')^! \circ (h')^*$, and $(h')^* = res(h^*)$ we conclude. $\square$