Exactly what the title says. Can you find some topological space $X\subset\mathbb R^2$ such that $\pi_1(X)\neq0$, but $\mathrm H_1(X,\mathbb Z)=0$?
I've been told that this paper shows that the fundamental group of any subspace of $\mathbb R^2$ has a torsion-free fundamental group, so at the very least, for such a space to exist there has to be some torsion-free group with trivial abelianization. I do not know if such a group exists.
Edit: It turns out that simple torsion-free groups exist, so there are torsion-free groups with trivial abelianization.
Best Answer
By the references in this answer, fundamental groups of subsets of the plane are residually free, and in particular, if $\pi_1(X)$ is nontrivial it surjects onto a nontrivial free group. Because free groups have nontrivial abelianization, we see that $\pi_1(X)$ surjects onto an abelian group, and hence $H_1(X) = \pi_1(X)^{\text{ab}}$ is nontrivial.