Proving that sheaf is constant

Question

Assume we have a sheaf $F \in Sh(X) $ of $R$ modules such that for two open sets $U_1,U_2 \in X; U_1 \cup U_2 =X $ ($U_1 \cap U_2 $ is not empty and connected) $F|_{U_i}$ is a constant sheaf. Does it mean that $F$ is a constant sheaf? (is assumption that $U_1 \cap U_2 $ is connected necessary?) | $$$$My approach is the following: for isomoprhism $R^{*}|_{U_i} \rightarrow F|_{U_i} $ (where $R^{*}$ is constant sheaf) we get a morphism (from adjunction) $R|_{U_i} \rightarrow F|_{U_i} $ (for R constant presheaf: $R(V) =R $) since it is isomorphism on stalks we get that for each $r \in R$ for each open $W \subset U_i $ there exist section $s \in F(W) $ such that for each $x \in W$ $s_x =r$( image in stalk at point x).$$$$ Now this gives us that identity function $ \bigcup_{x \in X} R_x \rightarrow \bigcup_{x \in X} R_x$ (where domain has topology coming from sheaf F, and codomain topology coming from constant sheaf - just like in sheaffication procedure) is in fact continuous hence we get a well defined morphism from $F$ to constant sheaf, which is isomorphism on stalks and thus this morphism is an isomoprhism in the category $Sh(X)$. is this the right approach? Is assumption that $U_1 \cap U_2 $ is connected necessary? Thank you for all your answers.

Answer 1

as i used this Lemma in my BA thesis i would advise you to read the proof of lemma 5.7.5 on Pierre Shapiras Lecture Notes Algebra and Topology from this link

https://webusers.imj-prg.fr/~pierre.schapira/lectnotes/AlTo.pdf

It covers exactly what you are looking for. And i just want to say that it is a necessary assumption that the intersection is connected.

Good Luck