$\DeclareMathOperator{\res}{res}$Let $X$ be a topological space, $p \in X$ a point, $U \subset X$ an open subset covered by $\bigcup_{i \in I}U_i$, and $S$ a set (or an abelian group). I'm trying to show that the skyscraper sheaf $i_pS$ given by $$ i_pS(U) = \begin{cases} S & \text{if } \, p \in U \\ \{ e \} & \text{else} \, \end{cases} $$ is indeed a sheaf. Here's what I've tried so far.
First I want to show the gluing axiom, so I take sections $a_i \in i_pS(U_i)$ such that $\res^{U_i}_{U_i \cap U_j} a_i = \res^{U_j}_{U_i \cap U_j} a_j$. We need for there to exist a section $a \in i_pS(U)$ such that $\res^U_{U_i}a = a_i$ for all $i \in I$. But here's my problem - since the sections are elements of the set (or abelian group) $S$, I'm not sure what the restriction of one of these sections actually is (as opposed to the clear nature of a restriction when we talk about, say, the sheaf of differentiable functions).
I tried to do it case-wise, so say $p \in U_i$ and $p \notin U_j$ (so $p \notin U_i \cap U_j$) for example. Then $a_j \in \{e\}$ so $a_j =e$, and by our assumption, $\res^{U_i}_{U_i \cap U_j} a_i = e$ which means that $a_i$ is $e$ on the part of $U_i$ which overlaps with $U_j$. Does this mean that $a_i$ must be $e$ on the whole of $U_i$? I feel not, because then by this logic I think we would end up with $i_pS(U)= \{e \}$. But then if the $a_i$ take different values on different parts of $U$, how does one glue them together to construct a valid $a$, and what does this $a$ even look like?
I haven't yet attempted the other axiom for sheaves, but I think I'd run into similar problems with the above reasoning. Thank you for any help.
For simplicity we look at sheaves of abelian groups.
You can describe $i_pS$ as the sheaf of functions $f : U \to S$ with the property that $f(u) = e$ for all $u \neq p$. The restriction is the usual restriction of functions.
Now let $f_i : U_i \to S$ be a family of sections of $i_pS$, which agree on intersections. Then they glue uniquely to a function $f : X \to S$ and obviously $f \in i_pS(X)$.