Lemma 81.10.5. In Situation 81.2.1 let $X, Y/B$ be good. Let $f : X \to Y$ be a flat morphism of relative dimension $r$.

1. Let $Z \subset Y$ be a closed subspace with $\dim _\delta (Z) \leq k$. Then we have $\dim _\delta (f^{-1}(Z)) \leq k + r$ and $[f^{-1}(Z)]_{k + r} = f^*[Z]_ k$ in $Z_{k + r}(X)$.

2. Let $\mathcal{F}$ be a coherent sheaf on $Y$ with $\dim _\delta (\text{Supp}(\mathcal{F})) \leq k$. Then we have $\dim _\delta (\text{Supp}(f^*\mathcal{F})) \leq k + r$ and

$f^*[{\mathcal F}]_ k = [f^*{\mathcal F}]_{k+r}$

in $Z_{k + r}(X)$.

Proof. Part (1) follows from part (2) by Lemma 81.6.3 and the fact that $f^*\mathcal{O}_ Z = \mathcal{O}_{f^{-1}(Z)}$.

Proof of (2). As $X$, $Y$ are locally Noetherian we may apply Cohomology of Spaces, Lemma 68.12.2 to see that $\mathcal{F}$ is of finite type, hence $f^*\mathcal{F}$ is of finite type (Modules on Sites, Lemma 18.23.4), hence $f^*\mathcal{F}$ is coherent (Cohomology of Spaces, Lemma 68.12.2 again). Thus the lemma makes sense. Let $W \subset Y$ be an integral closed subspace of $\delta$-dimension $k$, and let $W' \subset X$ be an integral closed subspace of dimension $k + r$ mapping into $W$ under $f$. We have to show that the coefficient $n$ of $[W']$ in $f^*[{\mathcal F}]_ k$ agrees with the coefficient $m$ of $[W']$ in $[f^*{\mathcal F}]_{k+r}$. We may choose a commutative diagram

$\xymatrix{ U \ar[d] \ar[r] & V \ar[d] \\ X \ar[r] & Y }$

where $U, V$ are schemes, the vertical arrows are étale, and there exist points $u \in U$, $v \in V$ such that $u \mapsto v$ and such that $u, v$ map to the generic points of $W', W$. (Details omitted.) Consider the stalk $M = (\mathcal{F}|_ V)_ v$ as an $\mathcal{O}_{V, v}$-module. (Note that $M$ has finite length by our dimension assumptions, but we actually do not need to verify this. See Lemma 81.4.4.) We have $(f^*\mathcal{F}|_ U)_ u = \mathcal{O}_{U, u} \otimes _{\mathcal{O}_{V, v}} M$. Thus we see that

$n = \text{length}_{\mathcal{O}_{U, u}} (\mathcal{O}_{U, u} \otimes _{\mathcal{O}_{V, v}} M) \quad \text{and} \quad m = \text{length}_{\mathcal{O}_{V, v}}(M) \text{length}_{\mathcal{O}_{V, v}}( \mathcal{O}_{U, u}/\mathfrak m_ v \mathcal{O}_{U, u})$

Thus the equality follows from Algebra, Lemma 10.52.13. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).