**Proof.**
Assume (2). Using the material in Section 107.19 we conclude that $\omega _{X_ s}^{\otimes m}$ is globally generated. However, if $C \subset X_ s$ is a rational tail, then $\deg (\omega _{X_ s}|_ C) < 0$ by Algebraic Curves, Lemma 53.22.2 hence $H^0(C, \omega _{X_ s}|_ C) = 0$ by Varieties, Lemma 33.43.12 which contradicts the fact that it is globally generated. This proves (1).

Assume (1). First assume that $g \geq 2$. Assumption (1) implies $\omega _{X_ s}^{\otimes m}$ is globally generated, see Algebraic Curves, Lemma 53.22.6. Moreover, we have

\[ \mathop{\mathrm{Hom}}\nolimits _{\kappa (s)}(H^1(X_ s, \omega _{X_ s}^{\otimes m}), \kappa (s)) = H^0(X_ s, \omega _{X_ s}^{\otimes 1 - m}) \]

by duality, see Algebraic Curves, Lemma 53.4.2. Since $\omega _{X_ s}^{\otimes m}$ is globally generated we find that the restriction to each irreducible component has nonegative degree. Hence the restriction of $\omega _{X_ s}^{\otimes 1 - m}$ to each irreducible component has nonpositive degree. Since $\deg (\omega _{X_ s}^{\otimes 1 - m}) = (1 - m)(2g - 2) < 0$ by Riemann-Roch (Algebraic Curves, Lemma 53.5.2) we conclude that the $H^0$ is zero by Varieties, Lemma 33.43.13. By cohomology and base change we conclude that

\[ E = Rf_*\omega _{X/S}^{\otimes m} \]

is a perfect complex whose formation commutes with arbitrary base change (Derived Categories of Spaces, Lemma 73.25.4). The vanishing proved above tells us that $E \otimes ^\mathbf {L} \kappa (s)$ is equal to $H^0(X_ s, \omega _{X_ s}^{\otimes m})$ placed in degree $0$. After shrinking $S$ we find $E = f_*\omega _{X/S}^{\otimes m}$ is a locally free $\mathcal{O}_ S$-module placed in degree $0$ (and its formation commutes with arbitrary base change as we've already said), see Derived Categories of Spaces, Lemma 73.26.5. The map $f^*f_*\omega _{X/S}^{\otimes m} \to \omega _{X/S}^{\otimes m}$ is surjective after restricting to $X_ s$. Thus it is surjective in an open neighbourhood of $X_ s$. Since $f$ is proper, this open neighbourhood contains $f^{-1}(U)$ for some open neighbourhood $U$ of $s$ in $S$.

Assume (1) and $g = 1$. By Algebraic Curves, Lemma 53.22.6 the assumption (1) means that $\omega _{X_ s}$ is isomorphic to $\mathcal{O}_{X_ s}$. If we can show that after shrinking $S$ the invertible sheaf $\omega _{X/S}$ because trivial, then we are done. We may assume $S$ is affine. After shrinking $S$ further, we can write

\[ Rf_*\mathcal{O}_ X = (\mathcal{O}_ S \xrightarrow {0} \mathcal{O}_ S) \]

sitting in degrees $0$ and $1$ compatibly with further base change, see Lemma 107.9.3. By duality this means that

\[ Rf_*\omega _{X/S} = (\mathcal{O}_ S \xrightarrow {0} \mathcal{O}_ S) \]

sitting in degrees $0$ and $1$, see Duality for Spaces, Lemma 84.3.3^{1}. In particular we obtain an isomorphism $\mathcal{O}_ S \to f_*\omega _{X/S}$ which is compatible with base change since formation of $Rf_*\omega _{X/S}$ is compatible with base change (see reference given above). By adjointness, we get a global section $\sigma \in \Gamma (X, \omega _{X/S})$. The restriction of this section to the fibre $X_ s$ is nonzero (a basis element in fact) and as $\omega _{X_ s}$ is trivial on the fibres, this section is nonwhere zero on $X_ s$. Thus it nowhere zero in an open neighbourhood of $X_ s$. Since $f$ is proper, this open neighbourhood contains $f^{-1}(U)$ for some open neighbourhood $U$ of $s$ in $S$.
$\square$

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