Various definitions of Order Embedding and Order Isomorphism From my textbook, Wikipedia, and other sources from the Internet, I have come across various definitions of Order Embedding and those of Order Isomorphism, so I'm very confused about them.
My questions:

*

*Are those definitions actually equivalent ?


*Are there a preferred definition for Order Embedding and that for Order Isomorphism?
Thank you so much!

Def$-1$ of Order Embedding:

Let $(A,\prec_A),(B,\prec_B)$ be posets. A mapping $f:A\to B$ is an order embedding from $A$ into $B$ if and only if the following condition holds: $$(1):\forall x,y\in A:x \preccurlyeq_A y\iff f(x) \preccurlyeq_B f(y)$$

Def$-2$ of Order Embedding:

Let $(A,\prec_A),(B,\prec_B)$ be posets. A mapping $f:A\to B$ is an order embedding from $A$ into $B$ if and only if both of the following conditions hold:
\begin{align}&(1): f\text{ is injective}\\&(2): \forall x,y\in A:x \prec_A y \iff f(x)\prec_B f(y)\end{align}

Def$-1$ of Order Isomorphism:

Let $(A,\prec_A),(B,\prec_B)$ be posets. A mapping $f:A\to B$ is an order isomorphism from $A$ into $B$ if and only if both of the following conditions hold:
\begin{align}&(1): f\text{ is surjective}\\&(2): \forall x,y\in A:x \preccurlyeq_A y\iff f(x)\preccurlyeq_B f(y)\end{align}

Def$-2$ of Order Isomorphism:

Let $(A,\prec_A),(B,\prec_B)$ be posets. A mapping $f:A\to B$ is an order isomorphism from $A$ into $B$ if and only if both of the following conditions hold:
\begin{align}&(1): f\text{ is bijective}\\&(2): \forall x,y\in A:x \prec_A y \iff f(x)\prec_B f(y)\end{align}

Def$-1$ of Order Isomorphism between tosets:

Let $(A,\prec_A),(B,\prec_B)$ be tosets. A mapping $f:A\to B$ is an order isomorphism from $A$ into $B$ if and only if both of the following conditions hold:
\begin{align}&(1): f\text{ is surjective}\\&(2): \forall x,y\in A:x \prec_A y \implies f(x)\prec_B f(y)\end{align}

Def$-2$ of Order Isomorphism between tosets:

Let $(A,\prec_A),(B,\prec_B)$ be tosets. A mapping $f:A\to B$ is an order isomorphism from $A$ into $B$ if and only if both of the following conditions hold:
\begin{align}&(1): f\text{ is bijective}\\&(2): \forall x,y\in A:x \preccurlyeq_A y \implies f(x)\preccurlyeq_B f(y)\end{align}

 A: Order embedding
$\mbox{def 1}\iff\mbox{def 2}.$
Note that the first definition implies injective:
Given $f:A\to B$ such that $\forall x,y\in A:x \preccurlyeq_A y\iff f(x) \preccurlyeq_B f(y)$.


*

*$f$ is injective:



$\forall x,y\in A \quad f(x)=f(y)\implies\begin{cases}f(x) \preccurlyeq_B f(y)\implies x \preccurlyeq_A y\\f(y) \preccurlyeq_B f(x)\implies y \preccurlyeq_A x\end{cases}{\mid}\implies x=y$

I am sure you can prove the rest.
I am not sure about "preferred definition" but when I first study this I learn with the second definition.

Order Isomorphism
Again we have $\mbox{def 1}\iff\mbox{def 2}.$
First note that def 1 can be written as surjective Order embedding which implies it is bijective. I am sure you can prove that the fact one use $\preccurlyeq$ and the other $\prec$ does not matter.
I am not sure about "preferred definition" but when I first study this I learn with the second definition.

Order Isomorphism between tosets is basically the same.
A: Notice that for order embedding 1 uses <= and 2 uses <.
Usually orders are reflexive.  Strict orders are
not reflexive.  < is a strict order, <= is not.
For example 5 <= 5 but not 5 < 5.  
The two definitions are equivalent.
Reflexive orders are prefered.  
