Pointed morphisms from smooth geometrically integral groups to abelian varieties are homomorphisms

First let $k$ be algebraically closed. The defect $c(x,y)=f(xy)-f(x)-f(y)$ is a morphism $G\times G\to A$, using [F1], and is zero on $G\times \{e_G\}$ and $\{e_G\}\times G$. Obtain $U$ from [F2]. Since $G\times G$ is dense open in the smooth integral $U\times G$, regard $c$ as a rational map on that product; [F1] extends it uniquely to $\bar{c}:U\times G\to A$. Its restriction to $U\times \{e_G\}$ is zero by generic agreement and separatedness. Apply the rigidity part of [F2] with rational base point $e_G\in G\subset U$: $\bar{c}(u,y)=\bar{c}(e_G,y)=0$ everywhere. Consequently $f(xy)=f(x)+f(y)$ as a scheme morphism identity.

For arbitrary $k$, extend scalars to an algebraic closure. Geometric integrality and smoothness of $G$ remain, so step 1.1 gives the required identity after extension. That identity holds already over $k$: the equalizer is closed, and its defining ideal sections vanish after faithful flat scalar extension, so are zero. Together with the assumed identity preservation, multiplication preservation also implies inverse preservation by the group inverse equations. Therefore $f$ is a homomorphism. AC is inherited from [F1]–[F2] and the algebraic closure.