
Re: Homomorphism of posets and lattices
Posted:
Sep 19, 2013 5:05 AM


On Wed, 18 Sep 2013, William Elliot wrote: > On Wed, 18 Sep 2013, quasi wrote: > > > If X,Y are posets, a function f:X > Y is called an order > > homomorphism if x <= y implies f(x) <= f(y). > > > > If X,Y are posets, a bijective function f:X > Y is called an > > order isomorphism if both f and f^(1) are order homomorphisms. > > > > Posets X,Y are said to be order isomorphic if there exists an > > order isomorphism f:X > Y. > > > > Questions: > > > > Let X,Y be posets and suppose f:X > Y and g:Y > X are > > order homomorphisms. > > > > (1) If f,g are both injective, must X,Y be order isomorphic? > > > No. X = Rx{0,1}; (a,b) <= (r,s) iff a <= r, b = s; Y = R > (x,0) > arctan x, (x,1) > pi + arctan x; y > (y,0). > > > (2) If f,g are both surjective, must X,Y be order isomorphic? > > No. X = R  (0,1); Y = R > x > x if x <= 0 > x > x  1 if 1 <= x > > y > y if y <= 0 > y > max{ 1,y } if 0 < y > > What happens if both are bijective? > Clearly, for all x,y, (x <= y > f(x) <= f(y)). If f(x) <= f(y): fgg^1(x) <= fgg^1(y)
gf(x) <= gf(y)

