reserve
  X for non empty set,
  FX for Filter of X,
  SFX for Subset-Family of X;

theorem
  for X,Y be non empty set, f be Function of X,Y,
  F be Filter of X holds f.:F is Filter of Y iff Y = rng f
  proof
    let X,Y be non empty set,f be Function of X,Y,
    F be Filter of X;
    hereby
      assume f.:F is Filter of Y;
      then Y in f.:F by CARD_FIL:5;
      then consider B being Subset of X such that
      B in F and
A1:   Y = f.:B by FUNCT_2:def 10;
      now
        let y be object;
        assume y in Y;
        then consider x being object such that
A2:     x in dom f and
        x in B and
A3:     y=f.x by A1,FUNCT_1:def 6;
        thus y in rng f by A2,A3,FUNCT_1:def 3;
      end;
      then Y c= rng f;
      hence Y=rng f; end;
      assume
A4:   Y = rng f;
      reconsider fF=f.:F as filter_base of Y by Th13;
A5:   fF c= <.fF.) by def3;
      <.fF.) c= fF
      proof
        let x be object;
        assume
A6:     x in <.fF.);
        then reconsider x1=x as Subset of Y;
        consider b2 be Element of fF such that
A7:     b2 c= x1 by A6,def3;
        consider bx being Subset of X such that
A8:     bx in F and
A9:     b2=f.:bx by FUNCT_2:def 10;
        reconsider fx1=f"(x1) as Subset of X;
A10:    dom f = X by FUNCT_2:def 1;
A11:    f"(f.:bx) c= f"(x1) by A7,A9,RELAT_1:143;
        bx c= f"(f.:bx) by A10,FUNCT_1:76;
        then
        bx c= f"(x1) by A11;
        then fx1 in F by A8,CARD_FIL:def 1;
        then f.:fx1 in fF by FUNCT_2:def 10;
        hence thesis by A4,FUNCT_1:77;
      end;
      then fF=<.fF.) by A5;
      hence thesis;
    end;
