Questions about the constant sheaf.

Question

Let $X$ be a topological space and $A$ an Abelian group. The constant sheaf $\mathcal{A}$ on $X$ determined by $A$ is defined as follows. Give $A$ the discrete topology and for any open set $U \subseteq X$, let $\mathcal{A}(U)$ be the ggroup of all continuous maps of $U$ into $A$. My question is why for any connected open set $U$, $\mathcal{A}(U) \cong A$? I think that since $A$ has the discrete topology, single points of $A$ are open. So the preimage $V$ of a point of $A$ under a contineous map $f$ is an open subset of $U$. Since $U$ is connected, $V=U$. Therefore every contineous map is determined by a point of $A$. Therefore $\mathcal{A}(U) \cong A$. Is this true? Thank you very much.