According to Hartshorne,

(Chapter, Ex 1.17): Let $X$ be a topological space, let $P$ be a point, and let $A$ be an abelian group. Define a sheaf $i_p(A)$ as follows: $i_P(A)(U)=A$ if $P\in U$ and $i_P(A)(U)=0$ $otherwise$ This Sheaf is called the Skyscraper Sheaf.

Show that this could be described as $i_*(A)$, where $A$ denotes the constant sheaf $A$ on the closed subspace ${\{P\}}^{-}$ and and $i:{\{P\}}^{-}\rightarrow X$ denotes the inclusion.

According to 'Stack Project':Let $x\in X$ and $i_x: \{x\}\rightarrow X$ be the inclusion map. Let $A$ be a set and think $A$ as a sheaf on the one point space $\{x\}$. We call $(i_x)_*(A)$ the skyscraper sheaf at $x$ with value A.

My Question: In one definition it is the push-forward of a Sheaf on a one-pointed space where as in the other definition it is the push-forward of not necessarily a one pointed space. Are both the definitions equivalent?


Yes, all definitions are equivalent. Here is why:

The closed subset $F=\overline{\{x\}}\stackrel {i}{\hookrightarrow}$ is irreducible (as the closure of an irreducible subset) and so the constant sheaf $i_F(A)$ on $F$ with fibre $A$ (which Hartshorne ambiguously denotes by just $A$) has value $\Gamma(V, i_F(A))$ on every non-empty subset $V\subset F$ open in $F$.
The key point now is that, from the definition of "closure", an open subset $U\subset X$ cuts $F$ if and only if it contains $P$ so that the value of both sheaves on $U$ in that case is: $$\Gamma(U,i_P(A))=\Gamma(U,i_*(i_F(A)))=A\quad (P\in U)$$ and the value of both sheaves on $U$ is in the other case, namely when $U$ is disjoint from $A$ or equivalently when $U$ doesn't contain $P$: $$\Gamma(U,i_P(A))=\Gamma(U,i_*(i_F(A)))=\{0\} \quad (P\notin U)$$ .

The definitions are indeed equivalent but the introduction of a constant sheaf on the closure of the singleton set $\{x\}$ looks like a useless and confusing complication.

  • $\begingroup$ Good answer. Just one question. Does it matter that $F$ is irreducible? $\endgroup$
    – Dog_69
    Nov 3 '20 at 10:15
  • $\begingroup$ @Dog-69 Yes it does: on an irreducible space the constant presheaf is already a sheaf and so the sections of the constant sheaf are trivial to describe. $\endgroup$ Nov 3 '20 at 14:03
  • $\begingroup$ OK, it was my fault. I thought the constant presheaf was already a sheaf, because it verifies all the axioms except when you look at "weird" coverings, which no one cares about usually, though in this case is vital. Thanks. $\endgroup$
    – Dog_69
    Nov 3 '20 at 16:17

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