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

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 $$\mathbb{R}^3$$?

• As described, without any identification specified on the third side of the triangle, this surface is a Möbius band, its fundamental group is $\mathbb Z$, and it is embeddable in $\mathbb R^3$. Are you sure this is the surface you intended to describe? – Lee Mosher Oct 11 at 18:56
• @LeeMosher Erm, no, I’m probably using this notation incorrectly because I’m not very familiar with it. The surface I’m trying to describe would be formed by gluing the two marked sides of the triangle in the indicated orientation so that the three points that appear as the triangle’s corners would actually be the same point. That’s not a möbius strip, is it? – Franklin Pezzuti Dyer Oct 11 at 19:16
• @LeeMosher Oh nevermind, I see my mistake. What I should have drawn was a “two-sided” disc with the two edges both marked with counterclockwise arrows. – Franklin Pezzuti Dyer Oct 11 at 19:18
• Yes, that's a projective plane, with fundamental group $\mathbb Z_2$. And its nonembeddability is addressed in the answer of @AnubhavMukherjee. – Lee Mosher Oct 11 at 21:15

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.