
theorem Th29:
  for S, T being non empty TopSpace, f being continuous Function
of S,T, a, b, c being Point of S, P being Path of a,b, Q being Path of b,c, P1
  being Path of f.a,f.b, Q1 being Path of f.b,f.c st a,b are_connected & b,c
  are_connected & P1 = f*P & Q1 = f*Q holds P1+Q1 = f*(P+Q)
proof
  let S, T be non empty TopSpace;
  let f be continuous Function of S,T;
  let a, b, c be Point of S;
  let P be Path of a,b;
  let Q be Path of b,c;
  let P1 be Path of f.a,f.b;
  let Q1 be Path of f.b,f.c;
  assume that
A1: a,b are_connected & b,c are_connected and
A2: P1 = f*P and
A3: Q1 = f*Q;
  for x being Point of I[01] holds (P1+Q1).x = (f*(P+Q)).x
  proof
    let x be Point of I[01];
A4: f.a,f.b are_connected & f.b,f.c are_connected by A1,Th23;
    per cases;
    suppose
A5:   x <= 1/2;
      then
A6:   2*x is Point of I[01] by BORSUK_6:3;
      thus (P1+Q1).x = P1.(2*x) by A4,A5,BORSUK_2:def 5
        .= f.(P.(2*x)) by A2,A6,FUNCT_2:15
        .= f.((P+Q).x) by A1,A5,BORSUK_2:def 5
        .= (f*(P+Q)).x by FUNCT_2:15;
    end;
    suppose
A7:   x >= 1/2;
      then
A8:   2*x-1 is Point of I[01] by BORSUK_6:4;
      thus (P1+Q1).x = Q1.(2*x-1) by A4,A7,BORSUK_2:def 5
        .= f.(Q.(2*x-1)) by A3,A8,FUNCT_2:15
        .= f.((P+Q).x) by A1,A7,BORSUK_2:def 5
        .= (f*(P+Q)).x by FUNCT_2:15;
    end;
  end;
  hence thesis;
end;
