The Stacks project

80.11 Push and pull

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

In this section we verify that proper pushforward and flat pullback are compatible when this makes sense. By the work we did above this is a consequence of cohomology and base change.

Lemma 80.11.1. In Situation 80.2.1 let

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

be a fibre product diagram of good algebraic spaces over $B$. Assume $f : X \to Y$ proper and $g : Y' \to Y$ flat of relative dimension $r$. Then also $f'$ is proper and $g'$ is flat of relative dimension $r$. For any $k$-cycle $\alpha $ on $X$ we have

\[ g^*f_*\alpha = f'_*(g')^*\alpha \]

in $Z_{k + r}(Y')$.

Proof. The assertion that $f'$ is proper follows from Morphisms of Spaces, Lemma 65.40.3. The assertion that $g'$ is flat of relative dimension $r$ follows from Morphisms of Spaces, Lemmas 65.34.3 and 65.30.4. It suffices to prove the equality of cycles when $\alpha = [W]$ for some integral closed subspace $W \subset X$ of $\delta $-dimension $k$. Note that in this case we have $\alpha = [\mathcal{O}_ W]_ k$, see Lemma 80.6.3. By Lemmas 80.8.3 and 80.10.5 it therefore suffices to show that $f'_*(g')^*\mathcal{O}_ W$ is isomorphic to $g^*f_*\mathcal{O}_ W$. This follows from cohomology and base change, see Cohomology of Spaces, Lemma 67.11.2. $\square$

Lemma 80.11.2. In Situation 80.2.1 let $X, Y/B$ be good. Let $f : X \to Y$ be a finite locally free morphism of degree $d$ (see Morphisms of Spaces, Definition 65.46.2). Then $f$ is both proper and flat of relative dimension $0$, and

\[ f_*f^*\alpha = d\alpha \]

for every $\alpha \in Z_ k(Y)$.

Proof. A finite locally free morphism is flat and finite by Morphisms of Spaces, Lemma 65.46.6, and a finite morphism is proper by Morphisms of Spaces, Lemma 65.45.9. We omit showing that a finite morphism has relative dimension $0$. Thus the formula makes sense. To prove it, let $Z \subset Y$ be an integral closed subscheme of $\delta $-dimension $k$. It suffices to prove the formula for $\alpha = [Z]$. Since the base change of a finite locally free morphism is finite locally free (Morphisms of Spaces, Lemma 65.46.5) we see that $f_*f^*\mathcal{O}_ Z$ is a finite locally free sheaf of rank $d$ on $Z$. Thus clearly $f_*f^*\mathcal{O}_ Z$ has length $d$ at the generic point of $Z$. Hence

\[ f_*f^*[Z] = f_*f^*[\mathcal{O}_ Z]_ k = [f_*f^*\mathcal{O}_ Z]_ k = d[Z] \]

where we have used Lemmas 80.10.5 and 80.8.3. $\square$

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