# $\mathsf{co\text{-}NP}$ and Cook reductions

Can someone help me understand the steps in this argument? There is a decision problem that is in $\mathsf{co\text{-}NP}$ (under standard Karp reductions) and is $\mathsf{NP}$-hard with respect to Cook reductions. Does this imply that if it is in $\mathsf{NP}$ then $\mathsf{NP} = \mathsf{co\text{-}NP}$ and if so, why?

Every NP-complete program $A$ is co-NP hard under Cook reductions: given a problem $B$ in co-NP, its complement $\overline{B}$ is in NP, so there is a polytime function $f$ such that $f(x) \in A$ iff $x \in \overline{B}$. Therefore the following is a Cook reduction from $B$ to $A$: given $x$, ask whether $f(x) \in A$, and return the opposite.