Does every simply connected, orientable, non-compact, 3-manifold embed in $\mathbb{R}^3$?

Question

Let $M$ be a simply connected, (orientable), non-compact, 3-manifold without boundary. Must $M$ be homeomorphic with a topological subspace of $\mathbb{R}^3$?

Answer 1

When the manifold is the universal cover of a compact $3$-manifold $M$ (to begin with, lets say without boundary) then you construct the embedding by hands, using geometrization. In your question let's replace $\Bbb R^3$ with $S^3$ and we will see the embedding into $\Bbb R^3$ comes for free in the situation you are interested in.

An instructive case comes from the case where your manifold is a connect-sum of lens spaces. This has been the subject of a few recent threads:

Universal covers of non-prime 3-manifolds

The basic idea is that you choose a collection of reducing spheres for the connect sum decomposition, call them $\Sigma$. Then $M \setminus \Sigma$ is a disjoint union of punctured lens spaces. Each of these have universal covers diffeomorphic to punctured spheres, so they embed in $S^3$. You then glue the embedded punctured spheres together (in $S^3$) so that the appropriate boundary spheres are glued together. If you view the embedded punctured spheres from the perspective of their complements in $S^3$ we are essentially doing the canonical construction of a Cantor set. There is the exceptional case of $\Bbb RP^3 \# \Bbb RP^3$ where we are constucting the standard embedding $S^0 \to S^3$.

But this is the basic idea. The remaining geometric 3-manifolds have universal covers that are also subsets of $S^3$, so you similarly glue these together along the (lifted) torus decomposition or sphere decompositions. The tori (being incompressible) will lift to copies of $\Bbb R^2$, so those gluings are a little easier to visualize.