Does $\# \mathsf{P}\subseteq \mathsf{FP}^{\mathsf{PH}}$? The Toda's theorem is a relationship between two different complexity classes: $ \# \mathsf{P} $ and  $PH$. He proved that $ \mathsf{PH}\subseteq \mathsf{P}^{\#\mathsf{P}} $.

I wonder the following statement holds.


  $ $    $\#\mathsf{P} \subseteq \mathsf{FP}^{\mathsf{PH}}$ 


If we prove this statement, $\# \mathsf{P}$ and $\mathsf{PH}$, that is, counting and alternation are $roughly$ $equivalent$ classes.
If your statement holds, then the polynomial hierarchy collapses: $\mathsf{\# P} \subseteq \mathsf{FP}^{\mathsf{PH}}$ iff $\mathsf{\# P} \subseteq \mathsf{FP}^{\mathsf{\Sigma_k P}}$ for some fixed $k$ (by definition). But then by Toda's Theorem we have $\mathsf{PH} \subseteq \mathsf{P}^{\mathsf{\# P}} \subseteq \mathsf{P}^{\mathsf{FP}^{\mathsf{\Sigma_k P}}} = \mathsf{P}^{\mathsf{\Sigma_k P}} = \mathsf{\Delta_{k+1} P}$.