Locally closed subspaces and sheaves. (Bredon page 11.) Let $A$ be a locally closed subspace of $X$ and let $\mathcal{B}$ be a sheaf on $A$. It is easily seen (since $A$ is locally closed) that there is a unique topology on the point set $\mathcal{B}\cup(X\times\{0\})$ such that $\mathcal{B}$ is a subspace and the projection onto $X$ is a local homeomorphism (we identify $A\times\{0\}$ with the zero section of $\mathcal{B}$).
Why does $A$ locally closed imply a unique topology above? In particular think of $A = \{(x,y) | y>0\}\cup\{(0,0)\} \subset \mathbb{R}^2$. Then $A$ is not locally closed. Are there several choices for a topology on $A$? If not, what kind of spaces will give me trouble?
 A: The problem with non-locally closed subsets is not that you will have more than one choice for the topology but rather that you will have none. Take $\mathcal{B}$ to the constant sheaf $\mathbb{Z}_A$ and assume there is some valid topology on $\mathcal{C} := \mathcal{B} \cup (X \times \{0\})$. Let $a$ be an arbitrary point of the subspace $A$, and pick a neighborhood $U$ of $a$ in $X$ and a section $s$ of the extension $\mathcal{C}$ such that for all $p \in U \cap A$ we have $s(p)=1$, this is possible since $\mathcal{C}|_A$ is the constant sheaf $\mathbb{Z}_A$. Now, we have that $U \cap (X \setminus A) = \{ t \in U : s(t)=0 \}$ which is open in $U$, proving that $A$ is locally closed.
A: Thanks to Omar's example illustrating the salient point, here's a proof of Bredon's statement.
Let $\mathcal C = \mathcal B \cup (X \times \{0\})$ as a point set. A topology on $\mathcal C$ is generated by the open sets of $\mathcal B$ together with all sets of the form $U\times\{0\}$ for $U\subseteq X$ open. Writing $\pi_\mathcal B:\mathcal B\to A$ and $\pi_0:X\times\{0\}\to X$ for the local homeomorphisms of the sheaves $\mathcal B$ and $X\times\{0\}$ respectively, we wish to define a local homeomorphism $\pi:\mathcal C\to X$. There are no problems with continuity of the group operations, so such a map will make $\mathcal C$ a sheaf.
Let of course
\begin{align}
\pi(c) =\begin{cases}
 \pi_{\mathcal{B}}(c) \quad &\text{if } c\in \mathcal{B}\\
 \pi_0(c) \quad &\text{if } c\in X\times\{0\}.
\end{cases}
\end{align}
Let $c\in\mathcal C$. If $c\notin\mathcal B$, then everything is OK, so assume $c\in\mathcal B$. Write $C$ for the neighborhood (in $\mathcal B$) on which $\pi_\mathcal{B}$ is a homeomorphism. Let $D$ be an $X$-open set such that $\pi_\mathcal{B}(C)=D\cap A$.
If we assume that $A$ is locally
closed, then there is a $X$-open neighborhood $U$ of $\pi(c)\in A$ such that
$U\cap(X\setminus A)$ is open in $U$. Thus so is $U\cap D\cap (X\setminus A)$, and we can write $U\cap D\cap(X\setminus
A)=U\cap D\cap V$ for some $X$-open set $V\subseteq X\setminus A$. Then $\pi_0^{-1}
(V) = V\times\{0\}$ is open in $\mathcal C$, and $\pi$ is a homeomorphism on the open neighborhood $(\pi_\mathcal{B})|_C^{-1}(U\cap D\cap A)\cup\pi_0^{-1}(V)$ of $c$.
