I want to proof the following set theory exercise (from C. Pinter)
Let $A$ and $B$ be partially ordered classes, and let $f : A → B$ be an increasing function; assume $\vec{f}(A)=B$. Prove that if $(L, U)$ is a cut of $B$, then $(f^{-1}(L), f^{-1}(U))$ is a cut of $A$.
I tried it by contradiction, but it came to nothing. I appreciate tips to proof it
For convenience let $L'=f^{-1}[L]$ and $U'=f^{-1}[U]$. It’s clear that $$L'\cup U'=f^{-1}[L]\cup f^{-1}[U]=f^{-1}[L\cup U]=f^{-1}[B]=A$$ and that $$L'\cap U'=f^{-1}[L]\cap f^{-1}[U]=f^{-1}[L\cap U]=f^{-1}[\varnothing]=\varnothing\,.$$ Thus, $\{L',U'\}$ is a partition of $A$.
Suppose that $a_0\le a_1\in L'$. Then $f(a_0)\le f(a_1)\in L$, so $f(a_0)\in L$, and therefore $a_0\in L'$, i.e., $L'$ is downward closed. A similar argument shows that if $a_0\le a_1$, and $a_0\in U'$, then $a_1\in U'$, so that $U'$ is upward closed. Thus, $\langle L',U'\rangle$ is a cut in $A$.