TITLE: Why tensor product of two sheaves of modules is a sheaf of modules?
QUESTION [11 upvotes]: Let $(X, \mathcal O_X)$ be  a ringed topological space. Consider two $\mathcal O_X$ modules, $\mathcal F$ and $\mathcal G$. First we define the tensor product presheaf $\mathcal F \otimes_{p,\mathcal O_X} \mathcal G$, which assigns every open set $U$ in $X$ the $\mathcal O_X(U)$-module $\mathcal F(U) \otimes_{\mathcal O_X(U)} \mathcal G(U)$. Now, $\mathcal F \otimes_{p,\mathcal O_X} \mathcal G$ is a presheaf of abelian groups. We take the the sheafication of the $\mathcal F \otimes_{p,\mathcal O_X} \mathcal G$ to obtain a sheaf of abelian groups, $\mathcal F \otimes_{\mathcal O_X} \mathcal G$. My question is for any open set $U$ in $X$ what is the $\mathcal O_X(U)$-module on   $(\mathcal F \otimes_{\mathcal O_X} \mathcal G)(U)$?

REPLY [19 votes]: If $(X,\mathcal O_X)$ is a scheme and if $U\subset X$ is is an open affine subscheme, then  we have for all quasi-coherent sheaves $\mathcal F,\mathcal G$ of $\mathcal O_X$-modules the extremely pleasant  equality of $\mathcal O_X(U)$-modules: 

$$(\mathcal F \otimes_{ \mathcal O_X} \mathcal G)(U)=      \mathcal F(U) \otimes_{\mathcal O_X(U)} \mathcal G(U)         $$ 

However if $U$ is not affine all hell can break loose:
For example if $X=\mathbb P^1_\mathbb C$ is the complex projective line and if $\mathcal F=\mathcal O_X(1), \mathcal G=\mathcal O_X(-1)$, then for these quasi-coherent sheaves $\mathcal F \otimes_{ \mathcal O_X} \mathcal G=\mathcal O_X$, so that for  $U=X$: 
$$(\mathcal F \otimes_{ \mathcal O_X} \mathcal G)(X)=  \mathcal O_X(X) =\mathbb C \neq  \mathcal F(X)\otimes _{\mathcal O_X(X)}\mathcal G(X)=\mathbb C^2\otimes_\mathbb  C  0=0$$    
Reference
The first displayed equality is proved in Qing-Liu, chapter 5, Proposition 1.12 (b), page 162.