Surface with fundamental group $\pi_1(A)\cong \mathbb{Z}_2$ embeddable in $\mathbb {R}^3$?

Question

It’s well known that the surface formed by gluing up opposite boundary points on a closed disc has fundamental group $\mathbb{Z_2}$.

However, it seems that surface can’t be realized in $\mathbb{R}^3$ without self-intersection (I think “embedded” is the proper terminology, but I don’t have a background in this stuff).

Is it possible to create a surface with fundamental group $\mathbb{Z_2}$ that can be embedded in $\mathbb{R}^3$?

Answer 1

Not possible. Alexander duality is an obstruction. Check Corollary 3.45 Hatcher. If $X\subset R^3$, then $H_1(X)$ has to be torsion-free. But $H_1(X)= \pi_1(X)= Z_2$ is torsion (and non-trivial) for your case.