The Stacks project

Proof. Observe that the morphism $b$ is a local complete intersection morphism by More on Algebra, Lemma 15.31.2 and hence the statement makes sense. Since $Z \to X$ is a regular immersion (and hence a fortiori quasi-regular) we see that $\mathcal{C}_{Z/X}$ is finite locally free and the map $\text{Sym}^*(\mathcal{C}_{Z/X}) \to \mathcal{C}_{Z/X, *}$ is an isomorphism, see Divisors, Lemma 31.21.5. Since $E = \text{Proj}(\mathcal{C}_{Z/X, *})$ we conclude that $E = \mathbf{P}(\mathcal{C}_{Z/X})$ is a projective space bundle over $Z$. Thus $E \to Z$ is smooth and certainly a local complete intersection morphism. Thus Lemma 42.59.10 applies and we see that

with $\mathcal{C}$ as in the statement there. Of course $\pi ^* = \pi ^!$ by Lemma 42.59.5. It remains to show that $\mathcal{F}$ is equal to the kernel $\mathcal{C}$ of the map $H^{-1}(j^*\mathop{N\! L}\nolimits _{X'/X}) \to H^{-1}(\mathop{N\! L}\nolimits _{E/Z})$.

Since $E \to Z$ is smooth we have $H^{-1}(\mathop{N\! L}\nolimits _{E/Z}) = 0$, see More on Morphisms, Lemma 37.13.7. Hence it suffices to show that $\mathcal{F}$ can be identified with $H^{-1}(j^*\mathop{N\! L}\nolimits _{X'/X})$. By More on Morphisms, Lemmas 37.13.11 and 37.13.9 we have an exact sequence

By the same lemmas applied to $E \to Z \to X$ we obtain an isomorphism $\pi ^*\mathcal{C}_{Z/X} = H^{-1}(\pi ^*\mathop{N\! L}\nolimits _{Z/X}) \to H^{-1}(\mathop{N\! L}\nolimits _{E/X})$. Thus we conclude. $\square$