The Stacks project

Lemma 6.29.3. In the situation described above, let $i \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{I})$ and let $\mathcal{G}$ be a sheaf on $X_ i$. For $U_ i \subset X_ i$ quasi-compact open we have

\[ p_ i^{-1}\mathcal{G}(p_ i^{-1}(U_ i)) = \mathop{\mathrm{colim}}\nolimits _{a : j \to i} f_ a^{-1}\mathcal{G}(f_ a^{-1}(U_ i)) \]

Proof. Let us prove the canonical map $\mathop{\mathrm{colim}}\nolimits _{a : j \to i} f_ a^{-1}\mathcal{G}(f_ a^{-1}(U_ i)) \to p_ i^{-1}\mathcal{G}(p_ i^{-1}(U_ i))$ is injective. Let $s, s'$ be sections of $f_ a^{-1}\mathcal{G}$ over $f_ a^{-1}(U_ i)$ for some $a : j \to i$. For $b : k \to j$ let $Z_ k \subset f_{a \circ b}^{-1}(U_ i)$ be the closed subset of points $x$ such that the image of $s$ and $s'$ in the stalk $(f_{a \circ b}^{-1}\mathcal{G})_ x$ are different. If $Z_ k$ is nonempty for all $b : k \to j$, then by Topology, Lemma 5.24.2 we see that $\mathop{\mathrm{lim}}\nolimits _{b : k \to j} Z_ k$ is nonempty too. Then for $x \in \mathop{\mathrm{lim}}\nolimits _{b : k \to j} Z_ k \subset X$ (observe that $\mathcal{I}/j \to \mathcal{I}$ is initial) we see that the image of $s$ and $s'$ in the stalk of $p_ i^{-1}\mathcal{G}$ at $x$ are different too since $(p_ i^{-1}\mathcal{G})_ x = (f_{b \circ a}^{-1}\mathcal{G})_{p_ k(x)}$ for all $b : k \to j$ as above. Thus if the images of $s$ and $s'$ in $p_ i^{-1}\mathcal{G}(p_ i^{-1}(U_ i))$ are the same, then $Z_ k$ is empty for some $b : k \to j$. This proves injectivity.

Surjectivity. Let $s$ be a section of $p_ i^{-1}\mathcal{G}$ over $p_ i^{-1}(U_ i)$. By Topology, Lemma 5.24.5 the set $p_ i^{-1}(U_ i)$ is a quasi-compact open of the spectral space $X$. By construction of the pullback sheaf, we can find an open covering $p_ i^{-1}(U_ i) = \bigcup _{l \in L} W_ l$, opens $V_{l, i} \subset X_ i$, sections $s_{l, i} \in \mathcal{G}(V_{l, i})$ such that $p_ i(W_ l) \subset V_{l, i}$ and $p_ i^{-1}s_{l, i}|_{W_ l} = s|_{W_ l}$. Because $X$ and $X_ i$ are spectral and $p_ i^{-1}(U_ i)$ is quasi-compact open, we may assume $L$ is finite and $W_ l$ and $V_{l, i}$ quasi-compact open for all $l$. Then we can apply Topology, Lemma 5.24.6 to find $a : j \to i$ and open covering $f_ a^{-1}(U_ i) = \bigcup _{l \in L} W_{l, j}$ by quasi-compact opens whose pullback to $X$ is the covering $p_ i^{-1}(U_ i) = \bigcup _{l \in L} W_ l$ and such that moreover $W_{l, j} \subset f_ a^{-1}(V_{l, i})$. Write $s_{l, j}$ the restriction of the pullback of $s_{l, i}$ by $f_ a$ to $W_{l, j}$. Then we see that $s_{l, j}$ and $s_{l', j}$ restrict to elements of $(f_ a^{-1}\mathcal{G})(W_{l, j} \cap W_{l', j})$ which pullback to the same element $(p_ i^{-1}\mathcal{G})(W_ l \cap W_{l'})$, namely, the restriction of $s$. Hence by injectivity, we can find $b : k \to j$ such that the sections $f_ b^{-1}s_{l, j}$ glue to a section over $f_{a \circ b}^{-1}(U_ i)$ as desired. $\square$


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 0A32. Beware of the difference between the letter 'O' and the digit '0'.