reducing #P-complete problem to NP problem

What would be the consequence and meaning of existence of polynomial reduction of #P-complete problem into NP problem (not NP-complete problem)?

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You have to be substantially more specific about what you mean by 'polynomial reduction' because #P problems aren't decision problems, so the usual notions of 'reduction' don't directly apply. Given that $P^{\#P}=PH$, though (for a suitable definition of $P^{\#P}$), it seems like one natural consequence of such a reduction would be the collapse of the polynomial hierarchy to at most $\Sigma_2$ ($=P^{NP}$). –  Steven Stadnicki Jul 11 '12 at 4:44
@Steven: $P^{\# P} \supseteq PH$ but equality is not known. –  sdcvvc Oct 16 '12 at 17:30
@sdcvvc D'oh, good catch - I'm not sure where I got the equality from. Though I think my conclusion still holds, regardless. –  Steven Stadnicki Oct 16 '12 at 18:26