Let $R$ be a ring and let $F$ and $G$ be sheaves of $R$-modules over a Hausdorff space $X$. Define the tensor product $F\otimes G$ as the sheafification of $U\mapsto F(U)\otimes G(U)$. (All tensor products over $R$)
Under which circumstances is it true that $$ F\otimes G (X)=F(X)\otimes G(X)? $$ Is this true if $X$ is simply connected or, say, contractible? We can also assume $R$ to be a field, if that helps.
Assuming $X$ simply connected/contractible won't help because those are merely conditions on the homotopy type, and the category of sheaves of $R$-modules is not a homotopy invariant.
Here's an example. Take $R$ a field, $X=S^1$, $F$ a nontrivial rank one local system and $G$ its dual. Then $H^0(X,F\otimes G)$ is nonzero but $H^0(X,F) \otimes H^0(X,G)$ vanishes. To get a contractible example replace $S^1$ with $\mathbb R^2$, and replace $F$ and $G$ with $i_\ast F, i_\ast G$, where $i:S^1 \to \mathbb R^2$ is the inclusion.
