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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!!

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  • $\begingroup$ What is the context? What is $N$? What is $Z$? What is $S$? What is $f$? $\endgroup$ – tomasz Jul 26 '13 at 20:25
  • $\begingroup$ Yes, they explicitly give the orientation. Read all the paragraphs on page 100 leading up to your quote, and the sentences after your quoted text. $\endgroup$ – Ryan Budney Jul 26 '13 at 20:37
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$T_z(Z)$ 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.

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