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Relativistic density-functional-theory calculations of Bi(111) thin films are performed to revisit their band structure and that of macroscopic samples. The band structure of a our 39-bilayer film ($\sim$~15~nm) shows that (1) $\sim$9-nm films are enough to describe that of Bi(111), (2) The two split surface-state metallic branches along the $\overline{ΓM}$ direction do not overlap with the bulk band at the zone boundary but lie within the A7-distortion-induced conduction-valence band gap, and (3) Neither the existence of the metallic surface states nor their observed splitting is related to inversion \emph{asymmetry}. Thus, the spin texture observed in such states is not caused by the lifting of the Kramers degeneracy and their splitting is not of the Rashba-type. We instead propose that (1) the large splitting of the metallic branches is a $m_j=\pm1/2$-$m_j=\pm3/2$ splitting and (2) the spin texture observed for the metallic branches may only occur because the almost unaltered strong covalent bonds retained by Bi(111) surface atoms cannot afford magnetic polarization. We emphasize that degeneracy at the $M$-point of the SBZ of Bi(111) -- implied by the translational symmetry of the surface -- is satisfied irrespectively of the presence of inversion symmetry centers. We show that the magnetic-moment discontinuity at $M$ does not exist, which also explains why the measured spin-polarization of the metallic branches vanishes near $M$. We induce the Rashba effect on the band structure of Bi(111) via different structural/electronic perturbations to reveal the actual lifting of the Kramers degeneracy and find that the magnitude of the perturbation imposed on a film correlates with the magnitude of the splitting and the localization of the Rashba-split states.