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For a point $x\in X$ and an abelian group $A$, the skyscraper sheaf $i_{\ast }A$, with $i:\{x\}\hookrightarrow X$, isRestrictions are identities or the map to zero, so it is a flasque sheaf. The global sections equal $A$ and higher cohomology vanishes:No closed-point assumption is needed for this calculation of skyscraper sheaf cohomology. On an arbitrary topological space, its nonzero stalks are at points of $\overline{\{x\}}$ when $A\ne 0$; only for a closed point is the nonzero stalk confined to $x$.

Choose the rational point $x=[1:0:\cdots :0]$ and the coherent sheaf $\mathcal{F}=i_{\ast }k$ supported there. By skyscraper sheaf cohomology, $H^0(X,\mathcal{F})=k$.

Its dual of a sheaf ${\mathcal{F}}^{\vee }=\mathcal{H}om(\mathcal{F},{\mathcal{O}}_X)$ is zero. Away from $x$ its source is zero. At $x$, a homomorphism $k\to {\mathcal{O}}_{X,x}$ must send one to an element annihilated by the maximal ideal. Since $n>0$, that ideal contains a nonzero coordinate parameter; the local ring is an integral domain, so the image must vanish. This proves that every local homomorphism vanishes. ConsequentlyThere is no contradiction with Serre duality: the ordinary sheaf dual suffices for locally free sheaves, but general coherent sheaves require an Ext functor. This example illustrates failure of ordinary sheaf-dual Serre duality for a skyscraper sheaf.

For a point inclusion $i:\{x\}\hookrightarrow X$ and an abelian group $A$, the skyscraper sheaf $i_{\ast }A$ has sections $A$ on opens containing $x$ and zero on other opens. Its restriction maps are identities or maps to zero. If $A\ne 0$, its nonzero stalks occur at every point of the closure $\overline{\{x\}}$: exactly those points whose every neighbourhood contains $x$. Thus the familiar assertion that only the stalk at $x$ is nonzero requires $x$ to be a closed point. It is always flasque, giving the stated skyscraper sheaf cohomology.