Remark 48.6.1. Consider a cartesian diagram

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

of quasi-compact and quasi-separated schemes with $(g, f)$ Tor independent. Let $V \subset Y$ and $V' \subset Y'$ be affine opens with $g(V') \subset V$. Form the cartesian diagrams

$\vcenter { \xymatrix{ U \ar[r] \ar[d] & X \ar[d] \\ V \ar[r] & Y } } \quad \text{and}\quad \vcenter { \xymatrix{ U' \ar[r] \ar[d] & X' \ar[d] \\ V' \ar[r] & Y' } }$

Assume (48.4.1.1) with respect to $K$ and the first diagram and (48.4.1.1) with respect to $Lg^*K$ and the second diagram are isomorphisms. Then the restriction of the base change map (48.5.0.1)

$L(g')^*a(K) \longrightarrow a'(Lg^*K)$

to $U'$ is isomorphic to the base change map (48.5.0.1) for $K|_ V$ and the cartesian diagram

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

This follows from the fact that (48.4.1.1) is a special case of the base change map (48.5.0.1) and that the base change maps compose correctly if we stack squares horizontally, see Lemma 48.5.2. Thus in order to check the base change map restricted to $U'$ is an isomorphism it suffices to work with the last diagram.

## 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Your email address will not be published.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 0E9S. Beware of the difference between the letter 'O' and the digit '0'.