The Stacks project

Proof. The problem is clearly local on $S$, hence we may replace $S$ by an affine open neighbourhood of $s$. The topology on $Y_ s$ is induced from the topology on $Y$, see Schemes, Lemma 26.18.5. Hence we can find a quasi-compact open $V \subset Y$ such that $V_ s = T$. The restriction of $h$ to $V$ is quasi-compact (as $S$ affine and $V$ quasi-compact), quasi-separated, locally of finite presentation, and flat hence flat of finite presentation. Thus after replacing $Y$ by $V$ we may assume, in addition to (1) and (2) that $Y_ s = T$ and $S$ affine.

Pick a closed point $y \in Y_ s$ such that $h$ is Cohen-Macaulay at $y$, see Lemma 37.22.7. By Lemma 37.23.4 there exists a diagram

such that $Z \to S$ is flat, locally of finite presentation, locally quasi-finite with $Z_ s = \{ y\}$. Apply Lemma 37.41.1 to find an elementary neighbourhood $(S', s') \to (S, s)$ and an open $Z' \subset Z_{S'} = S' \times _ S Z$ with $Z' \to S'$ finite with a unique point $z' \in Z'$ lying over $s$. Note that $Z' \to S'$ is also locally of finite presentation and flat (as an open of the base change of $Z \to S$), hence $Z' \to S'$ is finite locally free, see Morphisms, Lemma 29.48.2. Note that $Y_{S'} \to S'$ is flat and of finite presentation with geometrically reduced fibres as a base change of $h$. Also $Y_{s'} = Y_ s$ is geometrically connected. Apply Lemma 37.46.1 to $Z' \to Y_{S'}$ over $S'$ to get $V \subset Y_{S'}$ quasi-compact open satisfying (2) whose fibres over $S'$ are either empty or geometrically connected. As $V \to S'$ is open (Morphisms, Lemma 29.25.10), after replacing $S'$ by an affine open neighbourhood of $s'$ we may assume $V \to S'$ is surjective, whence (1) holds. $\square$