Universal structure-sheaf sections of an abelian scheme

Work over an affine open $V=\mathrm{Spec}R\subseteq S$, shrinking further so the complex $K$ of [F3] is finite free and concentrated in nonnegative degrees. At any $s\in V$, the geometric fibre $A_{\bar{s}}$ is smooth and connected, hence integral: regular local rings prevent its finitely many irreducible components from meeting, and connectedness leaves only one. Thus $A_s$ is geometrically integral and [F2] gives $H^0(A_s,\mathcal{O})=\kappa (s)$. The actual base-change map $H^0(K){\otimes }_R\kappa (s)\to H^0(K{\otimes }_R\kappa (s))=H^0(A_s,\mathcal{O})$ is surjective, because the global constant section $1$ maps to a basis of its target.

Apply the finite-free criterion in [F3] with $q=0$ to the map just proved surjective. Its preceding map in degree $-1$ is also surjective, since $K$ has no negative terms. The criterion consequently makes $H^0(K)$ finite locally free and gives $H^0(K){\otimes }_R{R}^{\mathrm{\prime }}\cong H^0(K{\otimes }_R{R}^{\mathrm{\prime }})$ for every $R$-algebra $R^{\mathrm{\prime }}$ locally near $s$. Its residue-field rank is one by step 1.1. Since $s$ was arbitrary, these neighbourhoods cover $S$, proving that $f_{\ast }{\mathcal{O}}_A$ is invertible and universally compatible with base change.

The unit map $u:{\mathcal{O}}_S\to f_{\ast }{\mathcal{O}}_A$ is a morphism of invertible sheaves which over each geometric fibre is an isomorphism (it sends $1$ to the constant function $1$); by [F4] it is an isomorphism, and the identity section $e:S\to A$ gives an inverse by pullback of functions, since $e^{\ast }u=\mathrm{id}$. The same argument applied to $A_T\to T$ and to arbitrary base change, including nonreduced $T$, gives ${\mathcal{O}}_T\cong f_{T,\ast }{\mathcal{O}}_{A_T}$ universally. This argument uses stalkwise and fibrewise isomorphisms supplied by the coherence theorem; it does not infer morphism equality from geometric points.