On the first step, both carbons of the C=C bond form C-Br bonds with one atom of the Br₂ molecule, pushing the second Br atom away with an additional pair of electrons as Br^-. It produces a bromonium ion.

On the second step, the Br^- formed on the first step, donates an electronn pair to the bromonium ion, forms a C-Br bond, and pushes the electron pair of the existing C-Br bond to Br, opening the three-membered ring.

On the second step, the Br^- approaches the bromonium cation from the opposite side, which makes the whole reaction an example of anti-addition.