Show that there exist nonorientable 1-dimensional manifolds if the Hausdorff condition is dropped from the definition of a manifold.
Solution
Theorem.
There exists a 1-dimensional manifold that is not orientable when
Hausdorffness is not required.
Proof by example.
Construct \[ X=\Bigl((\mathbb R\times\{+\})\sqcup(\mathbb R\times\{-\})\Bigr)\big/\!\!\sim \] with the equivalence relation: \[ (t,+)\sim(t,-)\ \text{for }t>0,\qquad (t,+)\sim(-t,-)\ \text{for }t<0, \] and no identification of \(0+\) with \(0-\).
Define the quotient map \(q\) and denote \(0_\pm=q(0,\pm)\) and \(U_\pm=q((-\varepsilon,\varepsilon)\times\{\pm\})\).
\(X\) is a 1-manifold.
- For \(t\neq0\) each point has a neighborhood inherited from one real line branch, so is homeomorphic to an interval.
- Near \(0_\pm\), the sets \(q([-\varepsilon,\varepsilon]\times\{\pm\})\) are intervals, since the quotient only folds the two copies together away from 0.
\(X\) is not Hausdorff. Any neighborhoods of \(0_+\) and \(0_-\) both contain points identified with small positive real numbers, hence intersect; thus no disjoint neighborhoods separate \(0_+\) and \(0_-\).
Put manifold charts \[\phi_+:U_+\to(-1,1),\ \phi_+(q(t,+))=t,\] \[\phi_-:U_-\to(-1,1),\ \phi_-(q(t,-))=t.\] On overlaps, for points coming from \(t<0\) the transition is orientation-preserving on one branch and orientation-reversing on the other branch. More precisely, the overlap map is locally \(t\mapsto t\) on positive coordinates and \(t\mapsto -t\) on negative coordinates. Hence no single chart orientation on \(U_-\) makes all overlap maps orientation-preserving.
Therefore \(X\) is nonorientable by definition of orientability for 1-manifolds via compatible orientations on overlaps.
So a non-Hausdorff 1-dimensional manifold can be nonorientable. ∎