is $2$–tight.

This notion gets weaker as $\kappa$ gets bigger. $f$ is $1$–tight iff $f(X)$ is scattered, so that “$2$–tight” is the first non-trivial case. $f$ is trivially $|X|^+$–tight. The usual projection from $[0,1]^2$ onto $[0,1]$ is not $2^{\aleph_0}$–tight.

Some easy equivalents to “$\kappa$–tight” are described in Lemma 2.2 of [@KU2]:

\[lemma-tight-equiv\] Assume that $X,Y$ are compact and $f : X \to Y$. Then $(1) \leftrightarrow (2)$. If $\kappa$ is finite and $Y$ is metric, then all four of the following are equivalent:

There is a loose family of size $\kappa$.

There is a disjoint family ${{\mathcal P}}$ of perfect subsets of $X$ with $|{{\mathcal P}}| = \kappa$ and a perfect $Q \subseteq Y$ such that $Q = f(P)$ for all $P \in {{\mathcal P}}$.

For some metric $M$ and $\varphi\in C(X,M)$, $\{y \in Y : |\varphi(f{^{-1}}\{y\})| \ge \kappa\}$ is uncountable.

Statement $(3)$, with $M = [0,1]$.

If $X,Y$ are both compact metric, then $f: X \to Y$ is $\kappa$–tight iff $\{y \in Y : |f{^{-1}}\{y\}| \ge \kappa\}$ is countable (see Theorem 2.7 of [@KU2]). Of course, the $\leftarrow$ direction is trivial. The $\rightarrow$ direction for non-metric $X$ and $\kappa = 2$ is refuted by the standard map from the 