According to Falconer's Book (Fractal Geometry: Mathematical Foundations and Applications), Proposition 14.3, we have the following result:
The Julia set $J=J(f)$ of $f$ is forward and backward invariant under $f$, so $J=f(J)=f^{-1}(J)$.
Proof: Let $z \in J$. Then $f^k(z) \not \rightarrow \infty$, and we may find $w_n \rightarrow z$ with $f^k\left(w_n\right) \rightarrow$ $\infty$ as $k \rightarrow \infty$ for all $n$. Thus, $f^k(f(z)) \not \rightarrow \infty$ and $f^k\left(f\left(w_n\right)\right) \rightarrow \infty$, where by continuity of $f, f\left(w_n\right)$ can be chosen as close as we like to $f(z)$. Thus $f(z) \in J$, so $f(J) \subset J$ which also implies $J \subset f^{-1}(f(J)) \subset f^{-1}(J)$.
Similarly, with $z$ and $w_n$ as above, if $f\left(z_0\right)=z$ then we may find $v_n \rightarrow z_0$ with $f\left(v_n\right)=w_n$, by the mapping properties of polynomials on $\mathbb{C}$. Hence, $f^k\left(z_0\right)=$ $f^{k-1}(z) \not \rightarrow \infty$ and $f^k\left(v_n\right)=f^{k-1}\left(w_n\right) \rightarrow \infty$ as $k \rightarrow \infty$, so $z_0 \in J$. Thus, $f^{-1}(J) \subset J$ which implies $J=f\left(f^{-1}(J)\right) \subset f(J)$.
I don't understand what the author means by the mapping properties of polynomials on $\mathbb{C}$. (Note that the book assumes that $f$ is always a complex polynomial defined on the whole of $\mathbb{C}$ and defines the Julia set as the boundary of the filled-in Julia set). Aside from that, the proof is okay, so I am just struggling with this step.
Thanks in advance for any help.