Proof.
Assume (2). Using the material in Section 109.19 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 33.44.12 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 33.44.13. 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 75.25.4). 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 75.26.5. 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 109.9.3. By duality this means that
Rf_*\omega _{X/S} = (\mathcal{O}_ S \xrightarrow {0} \mathcal{O}_ S)
sitting in degrees 0 and 11. 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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