# Why is the preimage orientation given by a transversal map smooth?

On page 100 of Differential Topology, Guillemin & Pollack define, given a smooth map $f: X \rightarrow Y$ between an orientedmanifold with boundary and an oriented boundaryless manifold and a submanifold $Z$ of $Y$ (also boundaryless and oriented) such that $f \pitchfork Z$ and $\partial f \pitchfork Z$, the preimage orientation for the manifold $S=f^{-1} (Z)$. What he does is first orient $df_{s}(N_s(S,X))$ (with $N_s(S,X)$ being the orthogonal complement in $T_s(X)$ of $T_s(S)$) so that

$df_s(N_s(S,X)) \oplus T_{f(s)}(Z)=T_{f(s)}(Y)$,

and then orient $N_s(S,X)$ and $S$ so that $df_s|_{N_s(S,X)}$ is an orientation preserving isomorphism and

$N_s(S,X) \oplus T_s(S) = T_s(X)$

My question is how can I show that this orientation is smooth (in the sense that around each $s \in S$ there is a smooth chart of S that preserves the orientation). I managed to do this in the case that $s \in \text{Int}(S)$ but my argument doesn't seem to translate at all to the case $s \in \partial S$.