## 80.10 Flat pullback

This section is the analogue of Chow Homology, Section 42.14.

Let $S$ be a scheme and let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $Z \subset Y$ be a closed subspace. In this chapter we will sometimes use the terminology *scheme theoretic inverse image* for the inverse image $f^{-1}(Z)$ of $Z$ constructed in Morphisms of Spaces, Definition 65.13.2. The scheme theoretic inverse image is the fibre product

\[ \xymatrix{ f^{-1}(Z) \ar[r] \ar[d] & X \ar[d] \\ Z \ar[r] & Y } \]

If $\mathcal{I} \subset \mathcal{O}_ Y$ is the quasi-coherent sheaf of ideals corresponding to $Z$ in $Y$, then $f^{-1}(\mathcal{I})\mathcal{O}_ X$ is the quasi-coherent sheaf of ideals corresponding to $f^{-1}(Z)$ in $X$.

Definition 80.10.1. In Situation 80.2.1 let $X, Y/B$ be good. Let $f : X \to Y$ be a morphism over $B$. Assume $f$ is flat of relative dimension $r$.

Let $Z \subset Y$ be an integral closed subspace of $\delta $-dimension $k$. We define $f^*[Z]$ to be the $(k+r)$-cycle on $X$ associated to the scheme theoretic inverse image

\[ f^*[Z] = [f^{-1}(Z)]_{k+r}. \]

This makes sense since $\dim _\delta (f^{-1}(Z)) = k + r$ by Lemma 80.9.1.

Let $\alpha = \sum n_ i [Z_ i]$ be a $k$-cycle on $Y$. The *flat pullback of $\alpha $ by $f$* is the sum

\[ f^* \alpha = \sum n_ i f^*[Z_ i] \]

where each $f^*[Z_ i]$ is defined as above. The sum is locally finite by Lemma 80.9.2.

We denote $f^* : Z_ k(Y) \to Z_{k + r}(X)$ the map of abelian groups so obtained.

An open immersion is flat. This is an important though trivial special case of a flat morphism. If $U \subset X$ is open then sometimes the pullback by $j : U \to X$ of a cycle is called the *restriction* of the cycle to $U$. Note that in this case the maps

\[ j^* : Z_ k(X) \longrightarrow Z_ k(U) \]

are all *surjective*. The reason is that given any integral closed subspace $Z' \subset U$, we can take the closure of $Z$ of $Z'$ in $X$ and think of it as a reduced closed subspace of $X$ (see Properties of Spaces, Definition 64.12.5). And clearly $Z \cap U = Z'$, in other words $j^*[Z] = [Z']$ whence the surjectivity. In fact a little bit more is true.

Lemma 80.10.2. In Situation 80.2.1 let $X/B$ be good. Let $U \subset X$ be an open subspace. Let $Y$ be the reduced closed subspace of $X$ with $|Y| = |X| \setminus |U|$ and denote $i : Y \to X$ the inclusion morphism. For every $k \in \mathbf{Z}$ the sequence

\[ \xymatrix{ Z_ k(Y) \ar[r]^{i_*} & Z_ k(X) \ar[r]^{j^*} & Z_ k(U) \ar[r] & 0 } \]

is an exact complex of abelian groups.

**Proof.**
Surjectivity of $j^*$ we saw above. First assume $X$ is quasi-compact. Then $Z_ k(X)$ is a free $\mathbf{Z}$-module with basis given by the elements $[Z]$ where $Z \subset X$ is integral closed of $\delta $-dimension $k$. Such a basis element maps either to the basis element $[Z \cap U]$ of $Z_ k(U)$ or to zero if $Z \subset Y$. Hence the lemma is clear in this case. The general case is similar and the proof is omitted.
$\square$

Lemma 80.10.3. In Situation 80.2.1 let $f : X \to Y$ be an étale morphism of good algebraic spaces over $B$. If $Z \subset Y$ is an integral closed subspace, then $f^*[Z] = \sum [Z']$ where the sum is over the irreducible components (Remark 80.5.1) of $f^{-1}(Z)$.

**Proof.**
The meaning of the lemma is that the coefficient of $[Z']$ is $1$. This follows from the fact that $f^{-1}(Z)$ is a reduced algebraic space because it is étale over the integral algebraic space $Z$.
$\square$

Lemma 80.10.4. In Situation 80.2.1 let $X, Y, Z/B$ be good. Let $f : X \to Y$ and $g : Y \to Z$ be flat morphisms of relative dimensions $r$ and $s$ over $B$. Then $g \circ f$ is flat of relative dimension $r + s$ and

\[ f^* \circ g^* = (g \circ f)^* \]

as maps $Z_ k(Z) \to Z_{k + r + s}(X)$.

**Proof.**
The composition is flat of relative dimension $r + s$ by Morphisms of Spaces, Lemmas 65.34.2 and 65.30.3. Suppose that

$A \subset Z$ is a closed integral subspace of $\delta $-dimension $k$,

$A' \subset Y$ is a closed integral subspace of $\delta $-dimension $k + s$ with $A' \subset g^{-1}(A)$, and

$A'' \subset Y$ is a closed integral subspace of $\delta $-dimension $k + s + r$ with $A'' \subset f^{-1}(W')$.

We have to show that the coefficient $n$ of $[A'']$ in $(g \circ f)^*[A]$ agrees with the coefficient $m$ of $[A'']$ in $f^*(g^*[A])$. We may choose a commutative diagram

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

where $U, V, W$ are schemes, the vertical arrows are étale, and there exist points $u \in U$, $v \in V$, $w \in W$ such that $u \mapsto v \mapsto w$ and such that $u, v, w$ map to the generic points of $A'', A', A$. (Details omitted.) Then we have flat local ring homorphisms $\mathcal{O}_{W, w} \to \mathcal{O}_{V, v}$, $\mathcal{O}_{V, v} \to \mathcal{O}_{U, u}$, and repeatedly using Lemma 80.4.1 we find

\[ n = \text{length}_{\mathcal{O}_{U, u}}( \mathcal{O}_{U, u}/\mathfrak m_ w\mathcal{O}_{U, u}) \]

and

\[ m = \text{length}_{\mathcal{O}_{V, v}}( \mathcal{O}_{V, v}/\mathfrak m_ w\mathcal{O}_{V, v}) \text{length}_{\mathcal{O}_{U, u}}( \mathcal{O}_{U, u}/\mathfrak m_ v\mathcal{O}_{U, u}) \]

Hence the equality follows from Algebra, Lemma 10.51.14.
$\square$

Lemma 80.10.5. In Situation 80.2.1 let $X, Y/B$ be good. Let $f : X \to Y$ be a flat morphism of relative dimension $r$.

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

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 80.6.3 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 67.12.2 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 67.12.2 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 80.4.4.) 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 10.51.13.
$\square$

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