If $x \in \Lambda$ then $f(x) \in \overline{J}$ by definition, and $f^n(f(x)) = f^{n+1}(x) \in \overline{J}$ for all $n \geq 0$, so $f(x) \in \Lambda$. This gives $f(\Lambda) \subset \Lambda$. Conversely, if $x \in \Lambda$ then since $F(K_0) = F(K_1) = J$, the surjectivity of $f \colon \overline{K_0} \cup \overline{K_1} \to \overline{J}$ guarantees there exists $y \in \overline{K_0} \cup \overline{K_1} \subset \overline{J}$ with $f(y) = x$. For every $k \geq 0$, $f^k(y) = f^{k-1}(x) \in \overline{J}$, so $y \in \Lambda$. Hence $\Lambda \subset f(\Lambda)$.