We show that if X is a space such that ${eta}QF(X)=QF({eta}X)$ and each stable $Z(X)^{sharp}$-ultrafilter has the countable intersection property, then there is a homeomorphism $h_X:vQF(X){ ightarrow}QF(vX)$ with $r_X={Phi}_{vX}{circ}h_X$. Moreover, if ${eta}QF(X)=QF({eta}X)$ and $vE(X)=E(vX)$ or $v{Lambda}(X)={Lambda}(vX)$, then $vQF(X)=QF(vX)$.