
theorem Th45:
  for T,S being TopSpace, f being Function of T,S
  holds f is open iff
    ex B being Basis of T st
      for V being Subset of T st V in B holds f.:V is open
proof
  let T,S be TopSpace, f be Function of T,S;
  hereby
    assume A1: f is open;
    reconsider B = the Basis of T as Basis of T;
    take B;
    let V be Subset of T;
    assume V in B;
    hence f.:V is open by A1, TOPS_2:def 1, T_0TOPSP:def 2;
  end;
  given B being Basis of T such that
    A2: for V being Subset of T st V in B holds f.:V is open;
  now
    let A be Subset of T;
    set Y = { G where G is Subset of T : G in B & G c= A };
    Y c= bool the carrier of T
    proof
      let g be object;
      assume g in Y;
      then ex G being Subset of T st g = G & G in B & G c= A;
      hence thesis;
    end;
    then reconsider Y as Subset-Family of T;
    set Z = { f.:G where G is Subset of T : G in Y };
    Z c= bool the carrier of S
    proof
      let h be object;
      assume h in Z;
      then ex G being Subset of T st
        h = f.:G & G in Y;
      hence thesis;
    end;
    then reconsider Z as Subset-Family of S;
    A7: for P being Subset of S holds P in Z implies P is open
    proof
      let P be Subset of S;
      assume P in Z;
      then consider G1 being Subset of T such that
        A5: P = f.:G1 & G1 in Y;
      ex G2 being Subset of T st
        G1 = G2 & G2 in B & G2 c= A by A5;
      hence thesis by A2, A5;
    end;
    assume A is open;
    then A = union Y by YELLOW_8:9;
    then f.:A = union Z by RELSET_2:14; :: f(UY) = Uf(Y)
    hence f.:A is open by A7, TOPS_2:19,TOPS_2:def 1;
  end;
  hence thesis by T_0TOPSP:def 2;
end;
