General projective hyperplanes decrease the defect of semismallness

The universal incidence $\mathcal{X}=\{(x,H):x\in X\cap H\}\subseteq X\times ({\mathbb{P}}^d{)}^{\vee }$ is a projective-space bundle of relative dimension $d-1$ over $X$: at each $x$ the hyperplanes containing $x$ are the linear forms vanishing at $x$, a rank-$d$ subbundle of the rank-$(d+1)$ bundle of linear forms, so the fibres are projective spaces of dimension $d-1$. Hence $\mathcal{X}$ is smooth, irreducible and of dimension $n+d-1$, and the fibre of $g$ at $(y,H)$ is $f^{-1}(y)\cap H$. A nonempty hyperplane cut of a projective variety of dimension $i+1$ has dimension $i$ unless the hyperplane contains a top-dimensional component, in which case its dimension is $i+1$; this is exactly [F4] applied component by component, taking the zero-dimensional components to be either contained in $H$ or disjoint from it.

Partition the universal locus of fibre dimension $i$ into $A_i=\{(y,H):y\in Y_i,\dim (f^{-1}(y)\cap H)=i\}$ and $B_i=\{(y,H):y\in Y_{i+1},\dim (f^{-1}(y)\cap H)=i\}$; both are constructible by [F1]. For fixed $y\in Y_i$ the fibre of $A_i$ over $y$ consists of the hyperplanes containing at least one of the finitely many top-dimensional components of $f^{-1}(y)$. Such a component spans a projective linear subspace of dimension at least $i$, so the hyperplanes containing it form a projective subspace of the dual of dimension at most $d-i-1$; hence every fibre of $A_i\to Y_i$ has dimension at most $d-i-1$, and [F2] applied to the finitely many components of $\overline{A_i}$ gives $\dim A_i\le \dim Y_i+d-i-1$. The projection $B_i\to Y_{i+1}$ has fibres of dimension at most $d$ because the hyperplane parameter is free, so $\dim B_i\le \dim Y_{i+1}+d$. Subtracting the source dimension $n+d-1$ of $g$ from twice the fibre dimension plus these parameter dimensions gives contributions at most $r(f)-i$ for $A_i$ and at most $r(f)-1$ for $B_i$; for $i=0$ the $A_0$ contribution is $\dim Y_0-n\le 0$ because $\dim Y_0\le \dim \overline{f(X)}\le n$ by [F2].

If $r(f)>0$, both bounds in step 2.1 are at most $r(f)-1$, including the separately treated $i=0$ case, so $r(g)\le r(f)-1$. If $r(f)=0$, the bounds are strictly negative for every $i>0$; for $i=0$ the contribution is zero, attained because $r(f)=0$ forces generic fibre dimension zero by [F2], so $g$ is generically finite on its image and has defect zero, while the strict bounds for positive fibre dimension prove smallness.

For the fixed-hyperplane claim, project each of the finitely many locally closed components of the $A_i$ and $B_i$ to $({\mathbb{P}}^d{)}^{\vee }$; a component whose image is not dense is absent from a general parameter fibre, and for a component with dense image [F2] bounds its general parameter fibre dimension by its dimension minus $d$. Intersecting the finitely many resulting nonempty parameter opens with the smooth-section open of [F3] gives a nonempty Zariski open; on it the same estimates as step 2.1 hold with $\dim A_i-d$, $\dim B_i-d$ and source dimension $n-1$ for $X_H$, so the identical arithmetic gives the fixed-hyperplane conclusions, For iteration, a smooth nonempty hyperplane section is pure of dimension one less by [F4], but need not be irreducible. The estimates of steps 1.1 and 2.1 apply unchanged to a smooth pure-dimensional finite union: all incidence components have the same source dimension, and [F2] is applied componentwise throughout. The defect is nonnegative, since on any irreducible component of dimension $m$ with generic fibre dimension $j$ its image has dimension $m-j$, so the corresponding full fibre locus contributes at least $j\ge 0$. Thus the positive-defect inequality applies at every successive smooth nonempty section, without an irreducibility assumption on those sections. Transversality of a prescribed finite algebraic Whitney stratification is the separately supplied condition named in the Statement, not a consequence of ordinary Bertini alone.