The Stacks project

Lemma 70.10.1. Let $f : X \to Y$ be a morphism of quasi-compact and quasi-separated algebraic spaces over $\mathbf{Z}$. Then there exists a direct set $I$ and an inverse system $(f_ i : X_ i \to Y_ i)$ of morphisms algebraic spaces over $I$, such that the transition morphisms $X_ i \to X_{i'}$ and $Y_ i \to Y_{i'}$ are affine, such that $X_ i$ and $Y_ i$ are quasi-separated and of finite type over $\mathbf{Z}$, and such that $(X \to Y) = \mathop{\mathrm{lim}}\nolimits (X_ i \to Y_ i)$.

Proof. Write $X = \mathop{\mathrm{lim}}\nolimits _{a \in A} X_ a$ and $Y = \mathop{\mathrm{lim}}\nolimits _{b \in B} Y_ b$ as in Proposition 70.8.1, i.e., with $X_ a$ and $Y_ b$ quasi-separated and of finite type over $\mathbf{Z}$ and with affine transition morphisms.

Fix $b \in B$. By Lemma 70.4.5 applied to $Y_ b$ and $X = \mathop{\mathrm{lim}}\nolimits X_ a$ over $\mathbf{Z}$ we find there exists an $a \in A$ and a morphism $f_{a, b} : X_ a \to Y_ b$ making the diagram

\[ \xymatrix{ X \ar[d] \ar[r] & Y \ar[d] \\ X_ a \ar[r] & Y_ b } \]

commute. Let $I$ be the set of triples $(a, b, f_{a, b})$ we obtain in this manner.

Let $(a, b, f_{a, b})$ and $(a', b', f_{a', b'})$ be in $I$. Let $b'' \leq \min (b, b')$. By Lemma 70.4.5 again, there exists an $a'' \geq \max (a, a')$ such that the compositions $X_{a''} \to X_ a \to Y_ b \to Y_{b''}$ and $X_{a''} \to X_{a'} \to Y_{b'} \to Y_{b''}$ are equal. We endow $I$ with the preorder

\[ (a, b, f_{a, b}) \geq (a', b', f_{a', b'}) \Leftrightarrow a \geq a',\ b \geq b',\text{ and } g_{b, b'} \circ f_{a, b} = f_{a', b'} \circ h_{a, a'} \]

where $h_{a, a'} : X_ a \to X_{a'}$ and $g_{b, b'} : Y_ b \to Y_{b'}$ are the transition morphisms. The remarks above show that $I$ is directed and that the maps $I \to A$, $(a, b, f_{a, b}) \mapsto a$ and $I \to B$, $(a, b, f_{a, b})$ are cofinal. If for $i = (a, b, f_{a, b})$ we set $X_ i = X_ a$, $Y_ i = Y_ b$, and $f_ i = f_{a, b}$, then we get an inverse system of morphisms over $I$ and we have

\[ \mathop{\mathrm{lim}}\nolimits _{i \in I} X_ i = \mathop{\mathrm{lim}}\nolimits _{a \in A} X_ a = X \quad \text{and}\quad \mathop{\mathrm{lim}}\nolimits _{i \in I} S_ i = \mathop{\mathrm{lim}}\nolimits _{b \in B} Y_ b = Y \]

by Categories, Lemma 4.17.4 (recall that limits over $I$ are really limits over the opposite category associated to $I$ and hence cofinal turns into initial). This finishes the proof. $\square$

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Your email address will not be published.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 0GS3. Beware of the difference between the letter 'O' and the digit '0'.