Is it true or false? Let $M$ a manifold of class $C^{\infty}$ not orientable and $p \in M$, then $M - \{p\}$ is orientable.

I believe this to be false, but I do not know a counterexample


If $M$ is the (boundaryless) Mobius strip and $p$ is a point that is not on the central loop, then $M\setminus\{p\}$ is not orientable, because the central loop has still got a neighbourhood which is diffeomorphic to the whole Mobius strip.

(actually, it's the same even if you pick $p$ on the central loop, though the diffeomorphism may be less obvious at first glance)


This is never true. That is, $M$ is orientable iff $M-\{p\}$ is orientable. This follows from the following characterization, which does not "see" discrete subsets:

Proposition. $M$ is not orientable iff there exists a finite sequence of charts $(U_i, \phi_i)_{1 \leq i \leq n}$ with $U_i$ connected, $n \geq 3$ and indices $\bmod n$ such that:

  • $U_i \cap U_{i+1} \neq \varnothing$ for all $i$
  • $\det(D(\phi_i \circ \phi_{i+1}^{-1}))>0$ for all $i$ except one
  • $\begingroup$ There is possibility of doing this is also normal fields, correct? $\endgroup$ – Croos Jun 29 '18 at 19:02
  • $\begingroup$ What do you mean with 'doing this' and 'normal fields'? $\endgroup$ – rabota Jun 29 '18 at 21:09

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