for $\pi$ of size $n+1$, with each $\pi(P_j) = Q$. Then each $\pi{\mathord {\upharpoonright}}P_j : P_j \to Q$ is $2$–supertight, $\ker (\pi{^{-1}}(Q)) \subseteq \bigcup_j P_j$, and $\pi{^{-1}}(Q)$ is $2$–superdissipated.

If tightness fails for $\pi{\mathord {\upharpoonright}}P_j$, then we could find uncountable closed $Q' \subseteq Q$ and disjoint closed $P_j^0, P_j^1 \subseteq P_j$ with $\pi(P_j^0) = \pi(P_j^1) = Q'$. If $P'_k = P_k \cap \pi{^{-1}}(Q')$, then the sets $P'_0, \ldots, P'_{j-1}, P_j^0, P_j^1, P'_{j+1}, \ldots, P'_n$ would be a loose family for $\pi$ of size $n+2$. If $\ker (\pi{^{-1}}(Q)) \not\subseteq \bigcup_j P_j$, we could find a perfect $R \subseteq \pi{^{-1}}(Q) \setminus \bigcup_j P_j$; then $\pi(R)$ is non-scattered (since all $\pi{^{-1}}\{y\}$ are scattered), and $R$ plus the $P_j$ would contradict the $(n+2)$–tightness of $\pi$. Finally, by Lemma \[lemma-ker-tight\], it is sufficient to prove that $\bigcup_j P_j$ is superdissipated, and this is done using the map into $Y \times \{0,1,\ldots,n\}$ which sends $x \in P_j$ to $(\pi(x),j)$.

Finally, we mention two lemmas for the case that $X$ does not contain a Cantor subset. $\pi: X \to Y$ is trivially $n$–supertight when all $|\pi{^{-1}}\{y\}| < n$, but also

\[lemma-lim-prop\] Assume that $\pi: X \to Y$, $X$ is compact, $Y$ is compact metric, each $|\pi{^{-1}}\{y\}| \le 