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.) enter image description here

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 $\partial (M-N)$.

  1. 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"


1 Answer 1


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).


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