## Return to Answer

2 fixed a statement

For $R^2$, connected open subspaces are precisely the non-compact orientable surfaces with genus zero (see this question for more info.)

For $R^3$, there is no homotopy criterion. The Whitehead manifold embeds in $R^3$ with complement the Whitehead continuum. The Whitehead manifold may be constructed as a union of solid tori, each one homotopically trivial in the next. One may modify the usual construction slightly by embedding each solid torus in the next with a little knot tied in it. Then one can show that this union does not embed in $R^3$. However, such manifolds are contractible, and therefore all homotopy invariants vanish.

To see that it doesn't embed in $R^3$ (or $S^3$), suppose you had an embedding in $S^3$. Then the complement of each solid torus is a knot complement, and each knot is a satellite of the next one. If each such knot were non-trivial, then you would have a knot whose complement contained infinitely many incompressible tori, contradicting Haken finiteness. So infinitely many of these solid tori must have trivial complement (this is exactly what you see in the embedding of the Whitehead manifold). But one may see that the little knot we tie in each embedding of a solid torus in the next must make the resulting knot complement have the little knot as a connect summand, giving a contradiction to embedding in $S^3$.

1

For $R^2$, connected open subspaces are precisely the orientable surfaces with genus zero (see this question for more info.)

For $R^3$, there is no homotopy criterion. The Whitehead manifold embeds in $R^3$ with complement the Whitehead continuum. The Whitehead manifold may be constructed as a union of solid tori, each one homotopically trivial in the next. One may modify the usual construction slightly by embedding each solid torus in the next with a little knot tied in it. Then one can show that this union does not embed in $R^3$. However, such manifolds are contractible, and therefore all homotopy invariants vanish.

To see that it doesn't embed in $R^3$ (or $S^3$), suppose you had an embedding in $S^3$. Then the complement of each solid torus is a knot complement, and each knot is a satellite of the next one. If each such knot were non-trivial, then you would have a knot whose complement contained infinitely many incompressible tori, contradicting Haken finiteness. So infinitely many of these solid tori must have trivial complement (this is exactly what you see in the embedding of the Whitehead manifold). But one may see that the little knot we tie in each embedding of a solid torus in the next must make the resulting knot complement have the little knot as a connect summand, giving a contradiction to embedding in $S^3$.
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