Lemma 42.59.10. Let $(S, \delta )$ be as in Situation 42.7.1. Consider a cartesian square

\[ \xymatrix{ X' \ar[d]_{f'} \ar[r]_{g'} & X \ar[d]^ f \\ Y' \ar[r]^ g & Y } \]

of schemes locally of finite type over $S$. Assume

both $f$ and $f'$ are local complete intersection morphisms, and

the gysin map exists for $f$

Then $\mathcal{C} = \mathop{\mathrm{Ker}}(H^{-1}((g')^*\mathop{N\! L}\nolimits _{X/Y}) \to H^{-1}(\mathop{N\! L}\nolimits _{X'/Y'}))$ is a finite locally free $\mathcal{O}_{X'}$-module, the gysin map exists for $f'$, and we have

\[ res(f^!) = c_{top}(\mathcal{C}^\vee ) \circ (f')^! \]

in $A^*(X' \to Y')$.

**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

\[ \xymatrix{ X' \ar[r] \ar[d] & P' \ar[r] \ar[d] & Y' \ar[d] \\ X \ar[r]^ i & P \ar[r]^ h & Y } \]

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

\[ \mathop{N\! L}\nolimits _{X/Y} = \left( \mathcal{C}_{X/P} \to i^*\Omega _{P/Y} \right) \]

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

\[ res(i^!) = c_{top}(\mathcal{C}^\vee ) \circ (i')^! \]

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.
$\square$

## Comments (2)

Comment #8872 by Eoin on

Comment #9214 by Stacks project on