A set-valued skyscraper sheaf and its stalks

If an open set $U$ does not contain $x_0$, then every section of ${\mathcal{S}}_{x_0,A}(U)$ is the unique element $\ast$, so locality and gluing are trivial. If $x_0\in U$ and $U={\bigcup }_i{U}_i$, then at least one $U_i$ contains $x_0$. Compatibility forces all sections on such $U_i$ to be the same element of $A$, and every $U_j$ not containing $x_0$ contributes only the unique section $\ast$. Thus there is a unique glued section on $U$. Therefore [L1] holds and ${\mathcal{S}}_{x_0,A}$ is a sheaf.

For the stalk at $x_0$, every neighbourhood of $x_0$ has section set $A$ and every transition map is the identity on $A$. Hence the colimit in [F1] is canonically $A$.

If $y$ has an open neighbourhood $V$ with $x_0\mathrm{\notin }V$, then every smaller neighbourhood of $y$ inside $V$ also has section set $\{\ast \}$. Hence the stalk diagram is eventually constant at $\{\ast \}$, so [F1] gives $({\mathcal{S}}_{x_0,A}{)}_y\cong \{\ast \}$.