The Stacks project

81.11 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 81.11.1. In Situation 81.2.1 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 66.40.3. The assertion that $g'$ is flat of relative dimension $r$ follows from Morphisms of Spaces, Lemmas 66.34.3 and 66.30.4. 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 81.6.3. By Lemmas 81.8.3 and 81.10.5 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 68.11.2. $\square$

Lemma 81.11.2. In Situation 81.2.1 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 66.46.2). 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 66.46.6, and a finite morphism is proper by Morphisms of Spaces, Lemma 66.45.9. 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 66.46.5) 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 81.10.5 and 81.8.3. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0EPC. Beware of the difference between the letter 'O' and the digit '0'.