I'm trying to prove trichotomy of well ordering and I have a question concerning the proof of the following fact: If $(A,\triangleleft_A)$ and $(B,\triangleleft_B )$ are two well ordered sets, then the following statements cannot be both true: $$ \textbf{1.} \ \exists a\in A \ \text{s.t.} \ pred(A,a,\triangleleft_A)\cong (B,\triangleleft_B),\\ \textbf{2.} \ \exists b\in B \ \text{s.t.} \ (A,\triangleleft_A) \cong pred(B,b,\triangleleft_B). $$ Likely wrong proof:
Suppose that $\textbf{1}$ and $\textbf{2}$ hold. Let us define the function $$g:pred(B,b,\triangleleft_B)\rightarrow pred(A,a,\triangleleft_A).$$ From $\textbf{1}$, since $\forall b'\in pred(B,b,\triangleleft_B)$, we have $g(b')=a'$ with $a'\in pred(A,a,\triangleleft_A)$, then $g$ is injective. In general it is not surjective because an element of $pred(A,a,\triangleleft_A)$ can be mapped in $B\setminus pred(B,b,\triangleleft_B)$.
But now, if $a''\in pred(A,a,\triangleleft_A)$ is such that $g^{-1}(a'')=b$, can we say that the function
$$
h:pred(B,b,\triangleleft_B)\rightarrow pred(A,a'',\triangleleft_A)
$$
is surjective and order preserving (taking advantage of the properties of $g$ and of the isomorphism in the hypothesis)? In such a case, since $h$ inherits the injectivity from $g$, it would be an isomorphism and, from $\textbf{2}$ we would get the contradiction:
$$
(A,\triangleleft_A) \cong pred(B,b,\triangleleft_B) \cong pred(A,a'',\triangleleft_A),
$$
that is, $(A,\triangleleft_A)$ would be isomorphic to an initial segment of itself.
I'm not able to prove that $h$ is indeed an isomorphism, maybe because it is not. Could you help me to find the mistake? (I hope this isn't too silly).
Thank you very much
First of all, it is always a good idea to fix witnesses (even when they are unique!). As (1) and (2) hold, let us fix order-isomorphisms $$\begin{gather} f : ( B , \triangleleft_B ) \to \mathrm{pred} ( A , a , \triangleleft_A ); \\ g : ( A , \triangleleft_A ) \to \mathrm{pred} ( B , b , \triangleleft_B ). \end{gather}$$ Consider the composition $gf : ( B , \triangleleft_B ) \to \mathrm{pred} ( B , b , \triangleleft_B )$. Note that $gf$ is an order-isomorphism between $( B , \triangleleft_B )$ and its range. It is relatively easy to show that the range of $gf$ is an initial segment of $( B , \triangleleft_B )$. Since the range of $gf$ cannot be $B$, there is some $b_0 \in B$ such that $\mathrm{pred} ( B , b_0 , \triangleleft_B )$ is the range of $gf$. It thus follows that $gf$ is an order-isomorphism between $( B , \triangleleft_B )$ and $\mathrm{pred} ( B , b_0 , \triangleleft_B )$, contradicting that no well-ordered set is order-isomorphic to a (proper) initial segment!