
Re: Order embedding
Posted:
Sep 16, 2013 11:32 AM


On Mon, 16 Sep 2013 14:24:06 +0300, Victor Porton <porton@narod.ru> wrote:
>William Elliot wrote: > >> Let X,Y be (partially) ordered sets. Are these definitions correct? >> >> f:X > Y is order preserving when >> for all x,y, (x <= y implies f(x) <= f(y). > >Yes. > >> f:X > Y is an order embedding when >> for all x,y, (x <= y iff f(x) <= f(y)). > >Yes.
I think no. Surely an embedding is required to be injective.
> >> f:X > Y is an order isomorphism when f is surjective >> and for all x,y, (x <= y iff f(x) <= f(y)). > >Yes.
No. An isomorphism must be bijective.
>> The following are immediate consequences. >> >> Order embedding maps and order isomorphisms are injections. > >Yes. > >> If f:X > Y is an order embedding, >> then f:X > f(X) is an order isomorphism. > >Yes. > >> Furthermore the composition of two order preserving, order >> embedding or order isomorphic maps is again resp., order > >> preserving, order embedding or order isomorphic. > >Yes. > >> Finally, the inverse of an order isomorphism is an order isomorphism. > >Yes. > >> That all is the basics of order maps, is it not? >> Or is the more to be included? > >Probably all. > >There are also Galois connections.

