  The neutron-shell structure of $^{25}$F was studied using quasi-free (p,2p) knockout reaction at 270A MeV in inverse kinematics. The sum of spectroscopic factors of $\pi$0d$_{5/2}$ orbital is found to be $1.0 \pm 0.3$. However, the spectroscopic factor for the ground-state to ground-state transition ($^{25}$F, $^{24}$O$_{g.s.}$) is only $0.36\pm 0.13$, and $^{24}$O excited states are produced from the 0d$_{5/2}$ proton knockout. The result shows that the $^{24}$O core of $^{25}$F nucleus significantly differs from a free $^{24}$O nucleus, and the core consists of 35% $^{24}$O$_{g.s}$. and 65% excited $^{24}$O. 
How different is the core of $^{25}$F from $^{24}$O$_{g.s.}$?