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I am sorry if this is a trivial question but I am a little confused right now so please bear with me.

Since non-orientable compact 2-manifolds without boundary cannot be embedded in three-dimensional Euclidean space is it true that all compact 2-manifolds without boundary embedded in three-dimensional Euclidean space are orientable?

Thanks everyone.

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While the proof in the smooth setting is easier, one also has the same conclusion in the topological setting:

Lemma. Let $M^{n-1}$ be a closed (compact and without boundary) connected topological manifold such that there exists $f: M\to R^n$, a topological embedding. (No assumption is made about $f(M)$ being a submanifold.) Then $M$ is orientable.

Proof. The proof is by applying twice the Poincare-Alexander duality to the subset $f(M)$ in $R^n$. First, apply it with $Z_2$ coefficients to conclude that $R^n\setminus f(M)$ has at least two components. Then apply duality again to conclude that $H^{n-1}(M; {\mathbb Z})$ is nonzero, i.e., is isomorphic to ${\mathbb Z}$. Therefore, $M$ has to be oriented. qed

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Yes, a smooth compact hypersurface (with no boundary) in $\Bbb R^n$ (more generally, in any simply connected manifold) is always orientable.

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  • $\begingroup$ The submanifold was not assumed to be smooth. Moreover, the image of embedding was not assumed to be a submanifold. $\endgroup$ – Moishe Kohan Jan 14 '14 at 6:28
  • $\begingroup$ @Ted Shifrin Thank you. That's what I was looking for. Also for anyone looking for a short proof please see here: Orientability of Hypersurface in R^n $\endgroup$ – orr Jan 14 '14 at 9:43

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