Endpoint boundaries are norm divisors

If the projection is not dominant, the image closure is a closed point $x$. Reducedness makes $W$ factor through $\{x\}\times {\mathbb{A}}^1$; since $\kappa (x)=k$, the only integral closed one-dimensional subscheme there is the whole affine line. Both endpoint lengths are one, so its boundary is zero.

Otherwise let $C$ be the reduced closure of $W$ in $T\times {\mathbb{P}}^1$. It is an integral curve, proper and dominant over $T$, with function field $L$ finite over $K$. Its intersection with $T\times {\mathbb{A}}^1$ is $W$. Admissibility implies $s\mathrm{\ne }0,1$ in $L$, so $g=(s-1)\mathrm{/}s\in L^{\times }$.

At $s=1$, $s$ is a unit and the order of $g$ is the length of the quotient by $s-1$; at $s=0$ it is minus the length of the quotient by $s$. Away from the endpoints in $W$, $g$ is a unit. At infinity, with $u=1\mathrm{/}s$, it equals $1-u$ and is a unit. Endpoint projection is an isomorphism onto $T$, so the residue-degree weights there are one. Consequently $p_{\ast }{\mathrm{div}}_C(g)=b[W]$ coefficient by coefficient.

Apply [F3] to $C\to T$ and $g$. Together with steps 1.1 and 2.1, and additivity of the divisor map, this proves both assertions.