Let \(\mcf\in \Sh(X)\) and \(s\in \mcf(U)\) be a section, and define \[ \supp s &\da \ts{p\in U \st s_p \neq 0} \subseteq U \\ \supp \mcf &\da \ts{p\in X \st \mcf_p\neq 0} \subseteq X ,\] where \(s_p\) denotes the germ of \(s\) in the stalk \(\mcf_p\). Show that \(\supp s\) is closed in \(U\) but \(\supp \mcf\) need not be closed in \(X\).
Solution
\(\supp(s)\) is closed.
By definition, \[ \supp(s) &\da \ts{p\in U \st \mcf\mid^U_p(s) \neq 0} \subseteq U \\ \implies \supp(s)^c \da U\sm \supp(s) &\da \ts{p\in U \st \mcf\mid^U_p(s) = 0} \subseteq U .\]
If \(p\in U\) and \(s=0\) in the stalk at \(p\), then \(s=0\) on an open neighborhood \(W_p\) of \(p\) with \(W_p \subseteq U\sm \supp(s)\), so every such \(p\) is interior. Hence \(U\sm\supp(s)\) is open in \(U\).
\(\supp(\mcf)\) need not be closed.
- Fix \(q\in X\) and an abelian group \(A\neq 0\). The skyscraper sheaf \(q_*\ul{A}\) is the pushforward of the constant sheaf on the point \(q\) along the inclusion \(q\injects X\):
Its sections and stalks are \[ q_* \underline{A}(U) = \begin{cases} A & q\in U \\ 0 & \text{else}, \end{cases} \qquad (q_* \underline{A})_p = \begin{cases} A & p\in\cl_X(\ts{q}) \\ 0 & \text{else}. \end{cases} \]
Exchange the roles of \(A\) and \(0\): \[ \mcf \da \left( U \mapsto \begin{cases} A & q\not\in U \\ 0 & q\in U \end{cases} \right)^{\scriptscriptstyle \mathrm{sh}} .\]
The presheaf and its sheafification \(\mcf\) have the same stalks, and \[ \mcf_p = \begin{cases} A & p\notin\cl_X(\ts{q}) \\ 0 & p\in\cl_X(\ts{q}). \end{cases} \]
Then for a fixed \(q\in X\), \(\supp \mcf = X \sm \cl_X(\ts{q})\).
- If there is some neighborhood \(U\ni p\) that does not contain \(q\), then the presheaf takes value \(A\) on every \(V \subseteq U\), so the colimit stabilizes and equals \(A\).
- If every neighborhood of \(p\) contains \(q\), then the presheaf takes value \(0\) on every neighborhood of \(p\), so the colimit is zero.
For example, take \(X \da \AA^1\slice{k}\) and \(q=0\); then \(\ts{0} = V(x)\) for \(x\in k[x]\) is closed, so \(\cl_X(\ts{0}) = \ts{0}\).
Thus \[ \supp \mcf = \AA^1 \sm \cl_X(\ts{0}) = \AA^1\smz = D(x) \] is open and not closed when \(k\) is infinite.