\text{~~~~[because $\rho {\circ}\rho = \rho$ and 
$\mu {\circ}\rho = \rho$]} \\
\mu  {\circ}f {\circ}\rho = \rho {\circ}f {\circ}\rho &\Rightarrow \\
\mu  {\circ}f {\circ}\rho = \mu {\circ}f {\circ}\mu &\end{aligned}$$ so that $\rho {\circ}f {\circ}\mu = \mu {\circ}f {\circ}\mu = \mu {\circ}f {\circ}\rho$.\
($\Leftarrow$) We have that: $$\begin{aligned}
\rho {\circ}f {\circ}\mu = \mu {\circ}f {\circ}\mu = \mu {\circ}f {\circ}\rho & \Rightarrow 
\text{~~~~[by applying $\rho$ to both sides]}
\\
\rho {\circ}\rho {\circ}f {\circ}\mu = \rho {\circ}\mu {\circ}f {\circ}\rho & \Rightarrow \text{~~~~[since $\rho {\circ}\rho = \rho$ and 
$\rho {\circ}\mu = \rho$]}\\
\rho {\circ}f {\circ}\mu = \rho {\circ}f {\circ}\rho & \Rightarrow\\
\mu {\circ}f {\circ}\mu = \rho {\circ}f {\circ}\rho. &\end{aligned}$$

Let us now observe that $\rho {\circ}f {\circ}\mu = \mu {\circ}f {\circ}\mu$: in fact, since $\rho = \rho {\circ}\mu$, this is equivalent to $\rho {\circ}\mu {\circ}f {\circ}\mu = \mu {\circ}f {\circ}\mu$, which is obviously equivalent to $\mu (f(\mu(C))) \subseteq \rho$.

Since $\rho = \mu {\circ}\rho$, we have that $\mu {\circ}f {\circ}\mu = \mu {\circ}f {\circ}\rho$ is equivalent to $\mu {\circ}(f {\circ}\mu) = \mu {\circ}(f{\circ}\mu) {\circ}\rho$. By the characterization of completeness in [@grs00 Lemma 4.2], since, by hypothesis, $f {\circ}\mu$ is continuous, we have that the completeness equation $\mu {\circ}(f {\circ}\mu) = \mu {\circ}(f{\circ}\mu) {\circ}\rho$ is equivalent 