ag.algebraic geometry - External tensor product of two (perverse) sheaves

So we have topological spaces $A,B$ and sheaves $F,G$ on $A,B$ of vector spaces over some fixed field $k$ and want to construct a sheaf $A otimes_k B$ on the product space $A times B$. You can write it down explicitly:

Let $W subseteq A times B$ be open. Then $(F otimes_k G)(W)$ consists of those elements $s in prod_{(a,b) in W} F_a otimes_k G_b$, such that for all $(a,b) in W$ there are open sets $a in U subseteq A, b in V subseteq B$ and $t in F(U) otimes_k G(V)$ such that $U times V subseteq W$ and for all $(c,d) in U times V$, we have $t_{c,d} = s_{c,d}$. Here $t mapsto t_{c,d}$ denotes the canonical map $F(U) otimes_k G(V) to F_c otimes_k G_d$.

Note that this obviously(!) yields a sheaf on $A times B$. On stalks, there is a canonical map $(F otimes_k G)_{a,b} to F_a otimes_k G_b$; a calculation shows that it is bijective. Remark that this agrees with the definition given by Strom Borman (the same universal property holds). But here you have a description of the sections of $F otimes_k G$. In particular, you see that if $F$ and $G$ are the sheaves of $mathbb{K}$-valued continuous functions on $A$ resp. $B$, then $F otimes_mathbb{K} G$ is a rather small subsheaf of the continuous functions on $A times B$.

The whole things makes more sense, when we take $A,B$ to be two $S$-schemes (or more generally, locally ringed spaces). Then we have the fibred product $A times_S B$ which can be constructed as above (I've written this up here (in german)). Here, the tensor product is the "right" sheaf.

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