# Oracle comparing $EXP$ with $UP$

Heller (Theorem 6) gave an oracle relative to which $$NP=EXP$$, and Homer & Selman gave an oracle relative to which $$P=UP$$ and $$\Sigma_2^P=EXP$$.

Beigel, Buhrman, Fortnow (freely available author's version) gave an oracle in which $$P=\oplus P$$ and $$NP=EXP$$ holds.

1. Is there an oracle that gives $$P\neq UP=coUP=NP=coNP=\oplus P=EXP$$?

2. Is there an oracle that gives $$P=UP=coUP\neq NP=coNP=\oplus P=EXP$$?

Or is it known $$UP\neq EXP$$ unconditionally?

$\mathsf{UP} \neq \mathsf{EXP}$ is open. A UP-generic oracle* should make $\mathsf{P} \neq \mathsf{UP} = \mathsf{EXP}$, and since $\mathsf{UP} \subseteq \mathsf{\oplus P} \subseteq \mathsf{EXP}$ relative to any oracle, this should resolve 1. (I say "should" because I haven't checked all the details...)
The quoted Beigel, Buhrman, and Fortnow paper gives a solution to 2 in Theorem 1.8: there is an oracle relative to which $\mathrm{P=Mod_3P}$ (which implies $\mathrm{P=UP}$), and $\mathrm{\oplus P=NP=EXP}$ (which, together with the first equality, actually implies $\mathrm{EXP=ZPP}$).
• In this case can it also hold $P\neq P^{UP}=NP$? – 1.. Dec 5 '20 at 18:44