I am sorry that my question might be stupid for experts, but I really do not know the answer.
Let $M$ be a smooth $n$-manifold. We assume that there exists a distance $d$ on $M$ such that $(M,d)$ is a length space and that the Hausdorff $Q$-measure associated to $d$ is Ahlfors $Q$-regular, which means the Hausdorff $Q$-measure of a ball $B(x,r)$ is comparable to $r^Q$.
My question is the following:
Suppose $M$ is simply connected. Let $E\subset M$ be a closed set. I wonder when will $M\backslash E$ be simply connected. In particular, I am interested in sufficient conditions on $E$ to ensure that $M\backslash E$ is simply connected in terms of Hausdorff dimension/measure.
Note that if $M$ is a simply connected domain in $\mathbb{R}^n$, then $M\backslash A$ is simply connected if $H^{n-2}(A)=0$, where $H^{n-2}$ is the Hausdorff $(n-2)$-measure associated to the standard Euclidean distance.
I am particularly interested in this result when $M$ is an equi-regular subRiemannian manifold. In the latter case, I know the result holds if $E$ has zero $Q$-capacity. I wonder what is the optimal condition on $E$ to ensure this in terms of Hausdorff measure/dimension.
