Solution
= Solution
For $F(t)=\prod_i(1-x_it)^{-1}$, 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>. Therefore
$$
\boxed{s_\lambda^F=s_\lambda(x).}
$$
For every <symmetric function> $u$, the same specialization gives $u^F=u(x)$.