I have a solution for this question:
Let $J$ and $E$ be well ordered sets; let $h:J→E$ . Show (i) implies (ii):
(i) $h$ is order preserving and its image is $E$ or a section of $E$ .
(ii) $h(α)=\min[E−h(S_α)]$ for all $α$.
(A section is defined for a well ordered set $X$ as: Given $a\in X$, let $S_a$ denote the set $S_a=\{x:x\in X \ \&\ x<a\}$. it is called section of $X$ by $a$.
The solution I have proves that $h(S_x)=S_{(h(x))}$ for all $x\in J$ and concludes that $h(\alpha)$ is the smallest element of $E$ not in $S_{h(α)}=h(S_α)$ . I can't understand how this conclusion is derived.
I mean i first thought that something like $h(A-B)=h(A)-h(B)$ might be useful, but then I found out that even if $A\subset B$, the image set $h(A-B)$ may not be equal to $h(A)-h(B)$. For example, if $h$ maps $A=\{1,2,3\}$ to $C=\{1,2\}$ as: $h(1)=h(2)=1$ and $h(3)=2$, then if $B=\{2,3\}$ then $h(A-B)={1}$ while $h(A)-h(B)=\phi$.
Let $e$ be the least element of $E\setminus h(S_\alpha)$. Since $h(\alpha)\not\in h(S_\alpha)$, $e\leq h(\alpha)$. If $e\neq h(\alpha)$, then $e<h(\alpha)$, so $e\in S_{h(\alpha)}=h(S_\alpha)$, which contradicts the definition of $e$. Thus we must have $e=h(\alpha)$.