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Topic: Homomorphism of posets and lattices
Replies: 16   Last Post: Sep 20, 2013 6:22 AM

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William Elliot

Posts: 1,241
Registered: 1/8/12
Re: Homomorphism of posets and lattices
Posted: Sep 19, 2013 5:05 AM
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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)



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