Homology 3-sphere smoothly embedded in $S^4$ disconnects $S^4$

Question

Suppose $M$ is a smoothly embedded homology 3-sphere in the unit 4-sphere $S^4$. Then is it true that $S^4-M$ has two components, i.e. $M$ separates $S^4$?

If $M=S^3$, then this is the Jordan curve theorem, but its proof uses that $S^3=D^3\cup D^3$, so the same proof cannot apply to homology 3-spheres.

According to p.11 of https://arxiv.org/pdf/2209.10735.pdf (right below Problem J), this seems to be true, but I can't see why.

Answer 1

This follows from Alexander duality. $$\tilde{H}_0(S^4 \setminus M) \cong \tilde{H}^3(M) \cong \Bbb Z$$

Thus $S^4 \setminus M$ has two components.