theorem
  for X1, X2 being non empty SubSpace of X st X = X1 union X2 for g
  being Function of X,Y holds g = (g|X1) union (g|X2)
proof
  let X1, X2 be non empty SubSpace of X such that
A1: X = X1 union X2;
  let g be Function of X,Y;
  reconsider h = g as Function of X1 union X2,Y by A1;
  X2 is SubSpace of X1 union X2 by TSEP_1:22;
  then
A2: h|X2 = g|X2 by Def5;
  X1 is SubSpace of X1 union X2 by TSEP_1:22;
  then h|X1 = g|X1 by Def5;
  hence thesis by A2,Th126;
end;
