 Solution

For $F(t)={\prod }_i(1-x_{i}t{)}^{-1}$, its coefficients are precisely the complete homogeneous symmetric functions $f_r=h_r(x)$. Thus the homomorphism sends every abstract generator $h_r$ to its ordinary value in the alphabet $x$. Equivalently the general product expansion is the ordinary Cauchy identity for symmetric functions. ThereforeFor every symmetric function $u$, the same specialization gives $u^F=u(x)$.