The Stacks project

Lemma 101.39.5. Assume given a $2$-commutative diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r]_-x \ar[dd]_ j & \mathcal{X} \ar[d]^ f \\ & \mathcal{Y} \ar[d]^ g \\ \mathop{\mathrm{Spec}}(A) \ar[r]^-z & \mathcal{Z} } \]

Choose a $2$-arrow $\gamma : z \circ j \to g \circ f \circ x$. Let $\mathcal{C}$ be the category of dotted arrows for the outer rectangle and $\gamma $. Let $\mathcal{C}'$ be the category of dotted arrows for the square

\[ \xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r]_-{f \circ x} \ar[d]_ j & \mathcal{Y} \ar[d]^ g \\ \mathop{\mathrm{Spec}}(A) \ar[r]^-z & \mathcal{Z} } \]

and $\gamma $. Then $\mathcal{C}$ is equivalent to a category $\mathcal{C}''$ which has the following property: there is a functor $\mathcal{C}'' \to \mathcal{C}'$ which turns $\mathcal{C}''$ into a category fibred in groupoids over $\mathcal{C}'$ and whose fibre categories are categories of dotted arrows for certain squares of the form

\[ \xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r]_-x \ar[d]_ j & \mathcal{X} \ar[d]^ f \\ \mathop{\mathrm{Spec}}(A) \ar[r]^-y & \mathcal{Y} } \]

and some choices of $y \circ j \to f \circ x$.

Proof. This lemma is a special case of Categories, Lemma 4.44.3. $\square$

Comments (0)

There are also:

  • 2 comment(s) on Section 101.39: Valuative criteria

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Your email address will not be published.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