The Engelking-Karlowicz theorem, and a useful corollary

Theorem (Engelking-Karlowicz, 1965). For cardinals $\kappa \le \lambda \le \mu \le 2^{\lambda }$, the following are equivalent:

1. ${\lambda }^{<\kappa }=\lambda$;
2. there exists a collection of functions, $⟨f_i:\mu \to \lambda \mid i<\lambda ⟩$, such that for every $X\in [\mu {]}^{<\kappa }$ and every function $f:X\to \lambda$, there exists some $i<\lambda$ with $f\subseteq f_i$.

Proof. (2)$\Rightarrow$(1) Suppose $⟨f_i:\mu \to \lambda \mid i<\lambda ⟩$ is a given collection. Then $$|\{f_i\upharpoonright \theta \mid i<\lambda ,\theta <\kappa \}|\le \lambda <{\lambda }^{<\kappa },$$
so there must exists some $f\in {}^{<\kappa }\lambda$ with $f⊈f_i$ for all $i<\lambda$.

(1)$\Rightarrow$(2) We follow a proof due to Shelah. Put:
$$W:=\{(a,\mathcal{A},g)\mid a\in [\lambda {]}^{<\kappa },\mathcal{A}\in [\mathcal{P}(a){]}^{<\kappa },g\in {}^{\mathcal{A}}\lambda \}.$$
Then $|W|={\lambda }^{<\kappa }=\lambda$, and we may fix an enumeration
$$W=\{(a_i,{\mathcal{A}}_i,g_i)\mid i<\lambda \}.$$
By $\mu \le 2^{\lambda }$, let $⟨B_{\alpha }\mid \alpha <\mu ⟩$ be a sequence of distinct subsets of $\lambda$.
For all $i<\lambda$, we now define $f_i:\mu \to \lambda$, by letting for all $\alpha <\mu$:
$$f_i(\alpha )=\{\begin{array}{ll}{g}_i(a_i\cap B_{\alpha }),& a_i\cap B_{\alpha }\in {\mathcal{A}}_i\\ 0,& \text{otherwise}\end{array}.$$
Finally, suppose that a set $X\in [\mu {]}^{<\kappa }$ and a function $f:X\to \lambda$ are given.
For all distinct $\alpha ,\beta \in X$, pick $x(\alpha ,\beta )\in B_{\alpha }\mathrm{\Delta }{B}_{\beta }$.
Put $a=\{x(\alpha ,\beta )\mid \alpha ,\beta \in X,\alpha \ne \beta \}$. Then, $|a|<\kappa$ and for all distinct $\alpha ,\beta \in X$, we have $a\cap B_{\alpha }\ne a\cap B_{\beta }$.
It follows that $|\mathcal{A}|=|a|$, where $\mathcal{A}:=\{a\cap B_{\alpha }\mid \alpha \in X\}$. It also follows that we may well-define a function $g:\mathcal{A}\to \lambda$ by letting:
$$g(a\cap B_{\alpha }):=f(\alpha ),(\alpha \in X).$$
Pick $i<\lambda$ such that $(a,\mathcal{A},g)=(a_i,{\mathcal{A}}_i,g_i)$. Then, $f\subseteq f_i$. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fb.svg">}$

Corollary (folklore). Assume CH.
Then there exists a function $G:[{\omega }_2{]}^{<{\omega }_1}\to {\omega }_1$ with the property that if $G(A)=G(B)$, then $A,B$ have the same order-type, and their intersection is an initial segment of $A$ and of $B$.
In particular, $G(A)=G(B)$ entails the existence of an order-preserving isomorphism $g:A\leftrightarrow B$, which is the identity on $A\cap B$.

Proof. For all $\alpha <{\omega }_2$, fix a bijection $g_{\alpha }:\alpha \leftrightarrow |\alpha |$. By ${\mathrm{\aleph }}_1^{{\mathrm{\aleph }}_0}={\mathrm{\aleph }}_1$ and the Engelking-Karlowicz theorem, we may also fix a sequence of functions $⟨f_i:{\mathrm{\aleph }}_2\to {\mathrm{\aleph }}_1\mid i<{\mathrm{\aleph }}_1⟩$ such that for every $X\in [{\mathrm{\aleph }}_2{]}^{<{\mathrm{\aleph }}_1}$ and $f:X\to {\mathrm{\aleph }}_1$, there exists some $i<{\mathrm{\aleph }}_1$ such that $f\subseteq f_i$.

We now define $G:[{\omega }_2{]}^{<{\omega }_1}\to {}^{<{\omega }_1}([{\omega }_1{]}^{<{\omega }_1})$, as follows.
Given $A\in [{\omega }_2{]}^{<{\omega }_1}$, let  $i<{\mathrm{\aleph }}_1$ be the least ordinal such that the order-preserving isomorphism ${\pi }_A:A\leftrightarrow \text{otp}(A)$ is extended by $f_i$, and put:
$$G(A):=\{(0,i)\}\cup ⟨g_{{\pi }_A^{-1}(j)}[A]\mid 0<j\le \text{otp}(A)⟩.$$

Next, suppose that $G(A)=G(B)$, for distinct $A,B\in [{\omega }_2{]}^{<{\omega }_1}$. It follows that:

• $\text{otp}(A)=\text{dom}(G(A))-1=\text{dom}(G(B))-1=\text{otp}(B)$;
• ${\pi }_A\cup {\pi }_B\subseteq f_i$, for $i:=G(A)(0)$;
• if $\alpha \in A\cap B$, then ${\pi }_A(\alpha )=f_i(\alpha )={\pi }_B(\alpha )$, and one of the following holds:
  

  $▸$ $f_i(\alpha )=0$, and $A\cap \alpha =\mathrm{\varnothing }=B\cap \alpha$;
  $▸$ $f_i(\alpha )>0$, and $A\cap \alpha =g_{\alpha }^{-1}[g_{\alpha }[A]]=g_{\alpha }^{-1}[G(A)({\pi }_A(\alpha ))]=g_{\alpha }^{-1}[G(B)({\pi }_B(\alpha ))]=g_{\alpha }^{-1}[g_{\alpha }[B]]=B\cap \alpha$.

So $A\cap B$ is an initial segment of $A$ and of $B$.

Let $g:={\pi }_B^{-1}\circ {\pi }_A$. Then $g:A\leftrightarrow B$ is an order-preserving isomorphism, and $g\upharpoonright A\cap B$ is the identity function, since ${\pi }_A\upharpoonright A\cap B={\pi }_B\upharpoonright A\cap B$ is simply the collapsing function from $A\cap B$ to $\text{otp}(A\cap B)$. $\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◻"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fb.svg">}$