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A sheaf is a bundle with some additional topological structure. Let $I$ be a topological space, with $\Theta$ its collection of open sets. A sheaf over $I$ is a pair $(A, p)$ where $A$ is a topological space $p : A \to I$ is a continuous map that is a local homeomorphism. This means that each point $x \in A$ has an open neighborhood $U$ in $A$ that is mapped homeomorphically by $p$ onto $p(U)$, and the latter is open in $I$. The category $\textbf{Top}(I)$ of sheaves over $I$ has such pairs $(A, p)$ a objects, and as arrows $k:(A, p) \to (B, q)$ the continuous maps $k : A \to B$ such that $q\circ k = p$ commutes. Such $k$ is in fact an open map (as is a local homeomorphism) and in particular $\text{Im} \ k = k(A)$ will be an open subset of $B$.
Here is my proof attempt.
A local homeomorphism $f: A \to I$ is an open map.
Proof. Let $U \subset A$ be open. Then for each $x \in U, \exists \ V_x \subset A$ which is open and such that $g = f|_{V_x} : V_x \simeq f(V_x)$ is a homeomorphism. Thus $V_x \cap U = $ an open set in $V_x$ and so $g(V_x \cap U) = g(V_x) \cap g(U)$ is open in $g(V_x)$.
Here I'm not sure whether unioning over $x$ would get us there.
Let's just focus on the local homeomorphism and I will make another question for the map $k$ if I need to.
Let $y\in f(U)$. You want to find an open neighborhood of $y$ which is contained inside $f(U)$. Let $x\in U$ with $f(x) = y$. Using the local homeomorphism property, there is an open neighborhood $V$ of $x$ such that $f|V$ is a homeomorphism of $V$ onto $f(V)$. Then $f|V$ is an open map; since $V \cap U$ is an open neighborhood of $x$ in $V$, then $f(V \cap U)$ is an open neighborhood of $y$ which is contained in $f(U)$.