Preimage orientation.

Question

On Guillemin and Pollack's Differential Topology Page 100.

Let $f: X \to Y$ be a smooth map with $f \pitchfork Z$ and $\partial f \pitchfork Z$, where $X,Y,Z$ are oriented and the last two are boundaryless. We define a preimage orientation on the manifold-with-boundary $S = f^{-1}(Z).$

But it never explicitly said what is preimage orientation. Is it $df_xN_x(S;X)$, where $N_x(S; X)$ be the orthogonal complement to $T_x(S)$ in $T_x(X)$?

Assume so, take the reflection map as an example. Consider $f: \mathbb{R} \to \mathbb{R}: f(x) = -x$, and let $Z = \{1\}.$ Hence $f$ is smooth with $f \pitchfork Z$ and $\partial f \pitchfork Z$, where $X,Y,Z$ are oriented, $Y$ is boundaryless, and grant $Z$ is boundaryless as well (see Is a single point boundaryless?).

Then the preimage orientation will be determined by $$df_xN_x(S;X) \oplus T_z(Z) = T_z(Y).$$

$T_z(Y)$ is just $\mathbb{R}$ with canonical orientation $+1$. Then what is the orientation of $T_z(Z)$? And how do they determine the orientation of $df_xN_x(S;X)$?

Very confused, thanks for your help!!

Answer 1

$T_z(Z)$ already has an orientation by hypothesis and the orientation of $df_x N_x(S;X)$ has the product orientation since the orientation of $T_z (Y)$ is also known.

Answer 2

One can make explicit Guillemin-Pollack definition of the orientation of a transversal inverse image. Let $x\in S$, then transversality says $T_xS=(d_xf)^{-1}(Z)$ and for any decomposition $T_xS\oplus E=T_xX$ (this $E$ could be the othogonal complement), the restriction $d_xf:E\to T_{f(x)}Y$ is injective, so induces an isomorphism onto its image $F=d_xf(E)$ and counting dimensions $T_{f(x)}Z\oplus F=T_{f(x)}Y$. Now we start with orientations in $X,Y$ and $Z$ to define one in $S$. Pick a positive basis $\{v_i\}$ of $T_{f(x)}Z$ and extend to a positive basis $\{v_i,w_j\}$ of $T_{f(x)}Y$, where $\{w_j\}$ is a basis of $F$. This determines uniquely a basis $\{\overline w_j\}$ of $E$ with $d_xf(\overline w_j)=w_j$. Now by definition, a basis $\{u_k\}$ of $T_xS$ is positive when $\{u_k,\overline w_j\}$ is a positive basis of $T_xX$.