 Solution

DefineCertainly $f(B)\subseteq B$. If $y\in B$, then $y\in f(X)$ and has a unique predecessor $x$ because $f$ is injective. For every $n$, writing $y=f^{n+1}(z)$ and using uniqueness gives $x=f^n(z)$; hence $x\in B$ and $y\in f(B)$. Thus $f(B)=B$.

If $f(a)\in B$, its unique predecessor must lie in $B$, so $a\in B$; consequently $f(A)\subseteq A$. If a point belonged to every $f^n(A)$ for $n\ge 1$, it would belong to $B$ and also to $f(A)\subseteq A$, which is impossible. Therefore