Subspace of $\Bbb R^2$ with fundamental group $\Bbb Z \oplus \Bbb Z$? [closed]

Does there exist a subspace of $\Bbb R^2$ with fundamental group $\Bbb Z \oplus \Bbb Z$?

• I don't think so. I think that connected subspace of $\mathbb R^2$ should retract on something like $U \backslash D$ where $D$ is a disjoint union of disks and $U$ is open and connected. Thus the fundamental group of subspace should be something like free group on $|D|$ generators. But I have no rigorous proof of what I am saying.
– user171326
Apr 8 '17 at 15:00
• Yeah intuitively that's what seems.... Apr 8 '17 at 15:01
• In general a connected subspace of $\mathbb{R}^2$ doesn't need to behave like that. The Hawaiian earring has a very big (and I think nonfree) fundamental group for example: en.wikipedia.org/wiki/Hawaiian_earring
– Jef
Apr 8 '17 at 15:12

No. I prove in this answer that $H_2(X;\Bbb Z)$ surjects onto $H_2(K(\pi_1 X, 1);\Bbb Z)$. (Actually, the coefficients can be whatever you want.) If you want to do this for gross $X$ this is still true but you need to use CW approximation. Further, for $X$ compact, it follows immediately from (Cech) Alexander duality that $H_2(X) =0$. If you further assume that $X$ is locally contractible, you can drop the Cech parenthetical; in this case, the techniques used in the proof can all be found in a first-year course on algebraic topology (even though they require heavier machinery than the fundamental group alone).
Unfortunately without these assumptions this gets more difficult. It is apparently a theorem of Zaslow that the higher homology of any subset of the plane vanishes identically. For $\pi_1 X = \Bbb Z^2$, $K(\Bbb Z^2,1) = T^2$, so we see that the second homology of any space with $\Bbb Z^2$ fundamental group is nontrivial.
• I would be interested in a proof specific to the case of $\Bbb Z^2$, perhaps that doesn't require much machinery. I'm skeptical it exists, though.