Does the Jordan-Brouwer Separation Theorem require smoothness and orientability of the hypersurface?

The Jordan-Brouwer Separation Theorem says that any connected compact hypersurface in Euclidean space divides the space into two connected components with the hypersurface their common boundary, cf. the lecture notes: http://www.math.uchicago.edu/~may/VIGRE/VIGRE2009/REUPapers/Schmaltz.pdf

However, in the notes https://www.jstor.org/stable/2323445 the author proved the theorem with two restrictions for oriented and smooth hypersurface. Although they are equivalent, say, any smooth hypersurface in Euclidean space is orientable, I don't know if this restriction can be dropped anyway. And the author said $C^1$- or $C^2$-smooth also satisfies the proof instead of $C^\infty$, but is $C^1$- or $C^2$-smoothness equivalent to orientability?

In summary, what are the necessary conditions for a hypersurface separates the Euclidean space?

A. connected

B. compact

C. $C^k$-smooth (and what $k$?)

D. oriented

Solution 1:

Suppose that $X\subset S^n$ is a compact subset which is homeomorphic to a topological manifold (possibly with boundary). Then Alexander Duality says: $$ \tilde{H}_0(S^n -X)\cong \check{H}^{n-1}(X)\cong H^{n-1}(X). $$ Therefore, if $X$ separates $S^n$, then $\tilde{H}_0(S^n -X)\ne 0$, hence, $dim(X)=n-1$ and at least one component of $X$ is orientable and has empty boundary. Conversely, if $X$ has codimension 1 in $S^n$ and has at least one component which is orientable and with empty boundary, then the same isomorphism says that the complement to $X$ is disconnected. If you want to get exactly two connected components, as in JST, then you have to assume that $X$ itself is connected, has empty boundary and is orientable.

Lastly, here is a proof that a closed (i.e. compact and with empty boundary) connected non-orientable $n-1$-dimensional manifold $X$ cannot be embedded in $S^n$:

Otherwise, $\tilde{H}_0(S^n-X, {\mathbb Z}_2)\cong H^{n-1}(X, {\mathbb Z}_2)\ne 0$, implying that $X$ separates $S^n$. But then, the argument above with integer (co)homology leads to a contradiction.