Assume (2). Using the material in Section \ref{section-relative-dualizing} 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 \ref{curves-lemma-rational-tail-negative} hence $H^0(C, \omega_{X_s}|_C) = 0$ by Varieties, Lemma \ref{varieties-lemma-check-invertible-sheaf-trivial} which contradicts the fact that it is globally generated. This proves (1). \medskip\noindent Assume (1). First assume that $g \geq 2$. Assumption (1)  implies $\omega_{X_s}^{\otimes m}$ is globally generated, see Algebraic Curves, Lemma \ref{curves-lemma-contracting-rational-tails}. Moreover, we have $$ \Hom_{\kappa(s)}(H^1(X_s, \omega_{X_s}^{\otimes m}), \kappa(s)) = H^0(X_s, \omega_{X_s}^{\otimes 1 - m}) $$ by duality, see Algebraic Curves, Lemma \ref{curves-lemma-duality-dim-1-CM}. 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 \ref{curves-lemma-rr}) we conclude that the $H^0$ is zero by Varieties, Lemma \ref{varieties-lemma-no-sections-dual-nef}. 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 \ref{spaces-perfect-lemma-flat-proper-perfect-direct-image-general}). 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 \ref{spaces-perfect-lemma-open-where-cohomology-in-degree-i-rank-r-geometric}. 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 $S$. \medskip\noindent Assume (1) and $g = 1$. By Algebraic Curves, Lemma \ref{curves-lemma-contracting-rational-tails} 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 \ref{lemma-genus}. By duality this means that $$ Rf_*\omega_{X/S} = (\mathcal{O}_S \xrightarrow{0} \mathcal{O}_S) $$ sitting in degrees $0$ and $1$\footnote{Use that $Rf_*\omega_{X/S}^\bullet = Rf_*R\SheafHom_{\mathcal{O}_X}(\mathcal{O}_X. \omega_{X/S}^\bullet) = R\SheafHom_{\mathcal{O}_S}(Rf_*\mathcal{O}_X, \mathcal{O}_S)$ by Duality for Spaces, Lemma \ref{spaces-duality-lemma-iso-on-RSheafHom} and Remark \ref{spaces-duality-remark-iso-on-RSheafHom} and then that $\omega_{X/S}^\bullet = \omega_{X/S}[1]$ by our definitions in Section \ref{section-relative-dualizing}.}. 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 $S$.