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When is the following map possible?

$A\boxtimes A(\mathcal{G}\times\mathcal{G})\rightarrow A(\mathcal{G}) \otimes A(\mathcal{G})$; where $\mathcal{G}$ is a group scheme, $A$ is a quasi coherent sheaf(of algebras) over $\mathcal{G}$ and $\boxtimes$ is the external tensor product of a sheaf given by $\pi_1^*A\otimes\pi_2^* A $.

In general, for any open set $U$, $A\boxtimes A(U\times U)\rightarrow A(U)\otimes A(U)$ doesnt hold true. What are the conditions required for this to hold? Will generation of $A$ by global sections suffice?

Please help.

share|cite|improve this question
    
There is a map in the other direction. Do you want to get a inverse of this? – Martin Brandenburg Oct 10 '10 at 22:02
    
Yes, I need a non-zero map. (isomorphism even better!) I know there is always a map in other direction. But when would the inverse(non-zero) map exist? I need to know the conditions that one require for this inverse to exist. Will generation of $\mathcal{A}$ by global sections be sufficient condition for this map to exist? – Neha Oct 11 '10 at 10:56
    
Let $(\mathcal{G},\mathcal{O})$ be an affine scheme, and $\mathcal{A}$ a quasi coherent sheaf over $\mathcal{G}$. Is $(\mathcal{A}\boxtimes \mathcal{A})(\mathcal{G}\mathcal{G})=\mathcal{A}(\mathcal{G})\otimes \mathcal{A}(\mathcal{G})$ ? – Neha Apr 22 '11 at 16:21
    
I meant $\mathcal{G}\times \mathcal{G}$ in the above statement. – Neha Apr 22 '11 at 16:22

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