n$, and $X$ has no Cantor subsets. Then $\pi$ is $n$–supertight.

If not, let $P_0, \ldots, P_{n-1} \subseteq X $ be a loose family, with each $\pi(P_j) = Q$. Then $Q$ has a Cantor subset, and each $P_j$ is homeomorphic (via $\pi$) to $Q$.

By the next lemma, the spaces $X$ we consider are always totally disconnected:

\[lemma-discon\] Assume that $\pi : X \to Y$, where $X$ is compact and $Y$ is metric. Assume that each $\pi{^{-1}}\{y\}$ is totally disconnected and $X$ does not contain a copy of the Cantor set. Then $X$ is totally disconnected.

Assume that $X$ is not totally disconnected. Fix a metric on $Y$ for which ${\mathrm{diam}}(Y) \le 1$. Obtain $K_s$ for $s \in 2^{<\omega}$ to satisfy:

$K_s$ is an infinite closed connected subset of $X$.

${\mathrm{diam}}(\pi(K_s)) \le 2^{- {\mathrm{lh}}(s)}$.

$K_{s{^{\mathord{\frown}}}0}, K_{s{^{\mathord{\frown}}}1} \subset K_s$ and $K_{s{^{\mathord{\frown}}}0} \cap K_{s{^{\mathord{\frown}}}1} = \emptyset$.

Assuming that this can be done, define $K_f = \bigcap_{n \in \omega} K_{f{\mathord {\upharpoonright}}n}$ for $f \in 2^\omega$. By (2), $|\pi(K_f)| = 1$; say $\pi(K_f) = \{y_f\}$. But $K_f$ is connected and $\pi{^{-1}}\{y_f\}$ is totally disconnected, so $|K_f| = 1$; say $K_f = \{x_f\}$. Then $f \mapsto x_f$ is a homeomorphism from $2^\omega$ into $X$, contradicting 