So I try to understand the following (which is taken from Dold, "Structure of the cobordism ring", Page 3/274, in the paragraph "1. La suite exacte de Wall."): https://eudml.org/doc/109581 ): Giving a compact smooth finite dimensional manifold $M$ we take the classifying map $f \colon M \to \mathbb{RP}^n$, such that $f^*(\gamma_{\infty})\cong \det(TM)$ (by compactness), where $\gamma_{\infty}$ is the universal line bundle over the infite-dimensional real projective space. If we homotopy $f$ in such a way that $f$ is transveral to $\mathbb{RP}^{n-1}$, so $f\pitchfork \mathbb{RP}^{n-1}$ and take the preimage of the new $f$, we get a submanifold $N=f^{-1}(\mathbb{RP}^{n-1})$ of $M$ of codimension 1 (Thom's transversality theorem). This submanifold $N$ ist orientable (as can be seen by a very short argument made in Stong ("Notes on Cobordism theory", Page 217 and 155)). The claim made from Dold and here (https://www.encyclopediaofmath.org/index.php/Orientation#General, in the paragraph beginning with "Along any path") is, that M-N is also orientable.
Ideas I had so far:
1.)
Is it true that $f|_{M-N}^*(\gamma_\infty)\cong \det (T (M-N))$? If that were the case we would have a classifying map into a contractible space $e_n$ (the diagram above is supposed to indicate the CW-Structure, and $e_n$ is the open n-cell), hence the result is a trivial bundle, so the first Stiefel-Whitney Class vanishes. Hence $M-N$ is orientable. The problem here is, that that doesn't seem to be the case. If it were true that $TM|_{M-N} = T(M-N)$ we again could conclude what we wanted, but the boundary $\partial (M-N)$ is tricky. Dold writes (Page 284, in the paragraph beginning with "Appendice : la suite exacte de Rokhlin.") that the boundary is diffeomorphic to two copies of $N$, but I would like to see it differently. In any way, we have a collar neighbourhood of $\partial (M-N)$, and $\partial (M-N)$ is a submanifold of $M$ of codimension 1. Meaning that the classyfing map at the collar neighbourhood of the surrounding manifold $\partial (M-N)$ is homotopic to the classifying map of $\partial (M-N)$.
- The other way is to use the orientation character (http://www.map.mpim-bonn.mpg.de/index.php?title=Orientation_character), but sadly this concept is new to me and I neither understood the train of thought laid out in Springer nor in Dold (both linked above). Is there any other way to tackle this? I am very grateful for any input leading into the right direction.
EDIT: I just realised the bottom right arrow should be an upward pointing "hook"
Here the answers to some of your questions.
Since taking determinant and pulling back commute (and the latter is natural) we indeed have for $i:M-\nu N \hookrightarrow M$ and $f:M \to RP^\infty$:
$$ detT(M-N) = det (i^*TM) = i^*detTM=i^*f^*\gamma_\infty=(fi)^*\gamma_\infty. $$
But you're right, $fi$ factors through a disk and hence gives the trivial line bundle which implies orientability.
To answer the question about the boundary $\partial M-\nu N$, note that we want to take out an open tubular neighborhood of $N$. Since $M$ and $N$ are both orientable, the normal bundle of $N$ is orientable and hence as a line bundle trivial. So we take out an open set diffeomorphic to $N \times (-1,1)$. Can you see why we remain with two copies of $N$ as boundary?
There are some other ways to see the orientability of $M-N$. Note that the equality of dimension here is crucial. You could use for example the orientation character (as you suggested) which says: $X$ is orientable iff it is orientable on every loop. That makes the problem pretty trivial. (By the way $\omega_1$ the first SW class is the orientation character, which you can pull back).