Questions about skyscraper sheaves. 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.
 A: 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.
