
theorem Th3:
  for X,Y being TopSpace st [#]X c= [#]Y & ex Xy being Subset of Y
  st Xy = [#]X & the topology of Y|Xy = the topology of X holds incl(X,Y) is
  embedding
proof
  let X,Y be TopSpace such that
A1: [#]X c= [#]Y;
A2: incl(X,Y) = id X by A1,YELLOW_9:def 1;
  set YY = Y| (rng incl(X,Y));
A3: [#]YY = rng incl(X,Y) by PRE_TOPC:def 5;
A4: [#]Y = {} implies [#]X = {} by A1;
  then reconsider h = incl(X,Y) as Function of X,YY by A3,FUNCT_2:6;
  set f = id X;
  given Xy being Subset of Y such that
A5: Xy = [#]X and
A6: the topology of Y|Xy = the topology of X;
A7: for P being Subset of X st P is open holds (h")"P is open
  proof
    let P be Subset of X;
    reconsider Q = P as Subset of YY by A2,A3;
    assume P is open;
    then
A8: P in the topology of X;
    (h")"P = h.:Q by A2,A3,TOPS_2:54
      .= Q by A2,FUNCT_1:92;
    hence (h")"P in the topology of YY by A5,A6,A2,A8;
  end;
A9: [#]X = {} implies [#]X = {};
A10: for P being Subset of YY st P is open holds h"P is open
  proof
    let P be Subset of YY;
    reconsider Q = P as Subset of X by A2,A3;
    assume P is open;
    then P in the topology of YY;
    then Q in the topology of X by A5,A6,A2;
    then Q is open;
    then f"Q is open by A9,TOPS_2:43;
    then f"Q in the topology of X;
    hence h"P in the topology of X by A1,YELLOW_9:def 1;
  end;
  take h;
  thus h = incl(X,Y);
  thus dom h = [#]X & rng h = [#]YY by A4,FUNCT_2:def 1,PRE_TOPC:def 5;
  thus h is one-to-one by A2;
  [#]YY = {} implies [#]X = {} by A2,A3;
  hence h is continuous by A10,TOPS_2:43;
  [#]X = {} implies [#]YY = {};
  hence thesis by A7,TOPS_2:43;
end;
