
theorem Th4:
for X,Y be non empty set, f be Function of X,Y, S be Field_Subset of X,
    M be Measure of S st f is bijective holds
  (ex G be Function of S,CopyField(f,S) st G = (.:f) |S
       & dom G = S & rng G = CopyField(f,S) & G is bijective)
& (ex F be Function of CopyField(f,S),S st F = ((.:f) |S)"
       & rng F = S & dom F = CopyField(f,S) & F is bijective)
proof
    let X,Y be non empty set, f be Function of X,Y, S be Field_Subset of X,
        M be Measure of S;
    assume
A1: f is bijective;

A2: dom ((.:f)|S) = S by FUNCT_2:def 1;
    rng ((.:f)|S) = .:f.:S by RELAT_1:115; then
A3: rng ((.:f)|S) = CopyField(f,S) by A1,Def1; then
    reconsider G = (.:f) |S as Function of S,CopyField(f,S) by A2,FUNCT_2:1;

A4: G is one-to-one by A1,FUNCT_1:52;

    G is onto by A3,FUNCT_2:def 3;
    hence
      ex G be Function of S,CopyField(f,S) st G = (.:f) |S
       & dom G = S & rng G = CopyField(f,S) & G is bijective by A2,A4,A3;

A5: dom (G") = CopyField(f,S) by A4,A3,FUNCT_1:33;
A6: rng (G") = S by A2,A4,FUNCT_1:33; then
    reconsider F = G" as Function of CopyField(f,S),S by A5,FUNCT_2:1;
    F is onto by A6,FUNCT_2:def 3;
    hence thesis by A4,A5,A6;
end;
