$z\in  \max((\mu {\circ}f {\circ}\mu)^{-1}(\downarrow\! y))$. Then, it turns out that $z \in \max(\{x\in \mu~|~\mu(f(x)) = \mu(f(\mu(z)))\})$. In fact, since $z=\mu(z)$, we have that $\mu(f(z))=\mu(f(\mu(z)))$. Moreover, if $u\in \{x\in \mu~|~\mu(f(x)) = \mu(f(\mu(z)))\}$ and $z \leq u$ then $\mu(f(\mu(u))) = \mu(f(u)) = 
\mu(f(\mu(z))) \leq y$, so that, by maximality of $z$, $z=u$, i.e., $z\in \max(\{x\in \mu~|~\mu(f(x)) = \mu(f(\mu(z)))\})$.\
$(\supseteq)$: Consider $y=\mu(f(\mu(w)))$ and $z\in \max(\{x\in \mu~|~\mu(f(x)) = y\})$. Then, $\mu(f(\mu(z))) = \mu(f(z)) = y$ so that $z\in (\mu {\circ}f {\circ}\mu)^{-1}(\downarrow\! y)$. If $u\in (\mu {\circ}f {\circ}\mu)^{-1}(\downarrow\! y)$ and $z \leq u$ then $\mu(f(\mu(z))) \leq \mu(f(\mu(u))) \leq y=\mu(f(\mu(w)))=\mu(f(\mu(z)))$. Hence, since $z\leq u \leq \mu(u)$ and by maximality of $z$, we have that $z=\mu(u)$, and in turn $z=u$. Thus, $z\in  \max((\mu {\circ}f {\circ}\mu)^{-1}(\downarrow\! y))$.

Consider sets of integers $\tuple{\wp(\mathbb{Z}),\subseteq}$ as concrete domain domain and the square operation ${\ensuremath{\mathit{sq}}}:\wp(\mathbb{Z}) \rightarrow \wp(\mathbb{Z})$ as concrete function, i.e., ${\ensuremath{\mathit{sq}}}(X) \triangleq \{x^2 ~|~ x\in X\}$, which is obviously additive and therefore continuous. Consider the abstract domain $\operatorname{Sign}\in \operatorname{Abs}(\wp(\mathbb{Z})_\subseteq)$, depicted in the following figure, that represents the sign of an integer variable.

(0,0) node\[name=1\] [[$\varnothing$]{}]{};

(-1,1) node\[name=2\] [[$\mathbb{Z}_{<0}$]{}]{}; (0,1) node\[name=3\] [[$0$]{}]{}; (1,1) node\[name=4\] [[$\mathbb{Z}_{>0}$]{}]{};

(-1,2) node\[name=5\] [[$\mathbb{Z}_{\leq 0}$]{}]{}; (0,2) node\[name=6\] [[$\mathbb{Z}_{\neq 0}$]{}]{}; (1,2) node\[name=7\] [[$\mathbb{Z}_{\geq 0}$]{}]{};

(0,3) node\[name=8\] [[$\mathbb{Z}$]{}]{}; (1) – 