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Let $X$ be a topological space and $P \in X$ a point. Let $A$ be an abelian group. The skyscraper sheaf $i_P(A)$ on $X$ is defined as follows: $i_P(A)(U) = A$ if $p \in U$ and $0$ otherwise. We need to verify that the stalk of $i_P(A)$ is $A$ at every point $Q \in \overline{\{P\}}$ and $0$ elsewhere, where $\overline{\{P\}}$ is the closure of the set consisting of the point $P$. I know that the stalk of $i_P(A)$ is $A$ at $Q \in X$ is \begin{align} {\lim_{\rightarrow}}_{Q \in U, U \text{ open}} i_P(A)(U) = \sqcup_{Q \in U, U \text{ open}} i_P(A)(U) / \sim. \qquad (1) \end{align} Let $Q \in \overline{\{P\}}$. Then the $Q$ is in very neighborhood of $P$. Let $U$ be an open neiborhood of $P$. Then $Q \in U$ and hence $i_P(A)(U) = A$. If $V$ is another open neiborhood of $P$, then $Q \in V$ and hence $i_P(A)(V) = A$. In (1), we need disjoint union $\sqcup$. But $i_P(A)(U) = A = i_P(A)(V)$. How can we verify that \begin{align} \sqcup_{Q \in U, U \text{ open}} i_P(A)(U) / \sim = A \end{align} for $Q \in \overline{\{P\}}$? Thank you very much.

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  • $\begingroup$ Think about the direct limit as gluing, where you identify things that are eventually equal. If $Q$ is in $\overline{\{P\}}$ then every neighborhood of $Q$ will contain $P$, and so ever neighborhood of $Q$ will have value $A$. $\endgroup$ Aug 29 '13 at 5:19
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I wouldn't use so much formulas. Words describe more properly and vividly what is going on. A sheaf is just a bunch of structures, associated to open subsets of a space, whose elements, called sections, can be glued together. The stalk of a sheaf at a point consists of the sections which are defined around that point, and two of these are identified when they agree on some small neighborhood.

Let $Q \in \overline{\{P\}}$. Then the $Q$ is in very neighborhood of $P$.

No, it's vice versa. If $Q \in \overline{\{P\}}$, then every open neighborhood of $Q$ also contains $P$ (by the way, this means that $Q$ is a specialization of $P$, a quote important notion in algebraic geometry). Hence, the open sets involved in computing the stalk at $Q$ all contain $P$. Therefore the group sections on each such open set is just $A$, and in the colimit you get $A$.

Conversely, if $Q \notin \overline{\{P\}}$, there is an open neighborhood $U$ of $Q$ which doesn't contain $P$. Now let us choose an element in the stalk at $Q$, say a section on some open neighborhood $V$ of $Q$. We can restrict it to $U \cap V$ because this doesn't change the element in the stalk. But $U \cap V$ doesn't contain $P$. Hence the section must be zero.

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  • $\begingroup$ thank you very much. We know that $ \overline{\{P\}} = \cap_{ P \in U, U \text{ is closed} } U$. If $Q \in \overline{\{P\}} = \cap_{ P \in U, U \text{ is closed} } U$, then $Q \in U$ for all $U$ which contains $P$. Therefore $Q$ is in every neiborhood of $P$. Is this true? I am confused. $\endgroup$
    – LJR
    Aug 29 '13 at 7:23
  • $\begingroup$ Yes, $Q \in \overline{P}$ means that $Q$ lies in every closed(!) set which contains $P$. But this means that $P$ lies in every open(!) set which contains $Q$. $\endgroup$ Aug 29 '13 at 7:37

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