Lemma 42.59.8. Let $(S, \delta )$ be as in Situation 42.7.1. Consider a cartesian diagram

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

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

1. $f$ is a local complete intersection morphism and the gysin map exists for $f$,

2. $X$, $X'$, $Y$, $Y'$ satisfy the equivalent conditions of Lemma 42.42.1,

3. for $x' \in X'$ with images $x$, $y'$, and $y$ in $X$, $Y'$, and $Y$ we have $n_{x'} - n_{y'} = n_ x - n_ y$ where $n_{x'}$, $n_ x$, $n_{y'}$, and $n_ y$ are as in the lemma, and

4. for every generic point $\xi \in X'$ the local ring $\mathcal{O}_{Y', f'(\xi )}$ is Cohen-Macaulay.

Then $f^![Y'] = [X']$ where $[Y']$ and $[X']$ are as in Remark 42.42.2.

Proof. Recall that $n_{x'}$ is the common value of $\delta (\xi )$ where $\xi$ is the generic point of an irreducible component passing through $x'$. Moreover, the functions $x' \mapsto n_{x'}$, $x \mapsto n_ x$, $y' \mapsto n_{y'}$, and $y \mapsto n_ y$ are locally constant. Let $X'_ n$, $X_ n$, $Y'_ n$, and $Y_ n$ be the open and closed subscheme of $X'$, $X$, $Y'$, and $Y$ where the function has value $n$. Recall that $[X'] = \sum [X'_ n]_ n$ and $[Y'] = \sum [Y'_ n]_ n$. Having said this, it is clear that to prove the lemma we may replace $X'$ by one of its connected components and $X$, $Y'$, $Y$ by the connected component that it maps into. Then we know that $X'$, $X$, $Y'$, and $Y$ are $\delta$-equidimensional in the sense that each irreducible component has the same $\delta$-dimension. Say $n'$, $n$, $m'$, and $m$ is this common value for $X'$, $X$, $Y'$, and $Y$. The last assumption means that $n' - m' = n - m$.

Choose a factorization $f = g \circ i$ where $i : X \to P$ is an immersion and $g : P \to Y$ is smooth. As $X$ is connected, we see that the relative dimension of $P \to Y$ at points of $i(X)$ is constant. Hence after replacing $P$ by an open neighbourhood of $i(X)$, we may assume that $P \to Y$ has constant relative dimension and $i : X \to P$ is a closed immersion. Denote $g' : Y' \times _ Y P \to Y'$ the base change of $g$ and denote $i' : X' \to Y' \times _ Y P$ the base change of $i$. It is clear that $g^*[Y] = [P]$ and $(g')^*[Y'] = [Y' \times _ Y P]$. Finally, if $\xi ' \in X'$ is a generic point, then $\mathcal{O}_{Y' \times _ Y P, i'(\xi )}$ is Cohen-Macaulay. Namely, the local ring map $\mathcal{O}_{Y', f'(\xi )} \to \mathcal{O}_{Y' \times _ Y P, i'(\xi )}$ is flat with regular fibre (see Algebra, Section 10.142), a regular local ring is Cohen-Macaulay (Algebra, Lemma 10.106.3), $\mathcal{O}_{Y', f'(\xi )}$ is Cohen-Macaulay by assumption (4) and we get what we want from Algebra, Lemma 10.163.3. Thus we reduce to the case discussed in the next paragraph.

Assume $f$ is a regular closed immersion and $X'$, $X$, $Y'$, and $Y$ are $\delta$-equidimensional of $\delta$-dimensions $n'$, $n$, $m'$, and $m$ and $m' - n' = m - n$. In this case we obtain the result immediately from Lemma 42.54.6. $\square$

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