Suppose $M$ is a connected manifold of dimension $n > 1$ and $B \subseteq M$ is a regular coordinate ball, is $M \setminus B$ connected?
I think the answer to this is definitely yes, but I'm currently trying to prove it and I'm not entirely sure how I can prove it. Since path-connectedness and connectedness are the same thing on manifolds, I was trying to prove it using path-connectedness as that's easier to prove than connectedness generally.
I was thinking of assuming of picking two points $p, q \in M \setminus B \subseteq M$, such that they had a path $\gamma : [0, 1] \to M$ connecting them which passed through $B$ and showing that it gave rise to a path $\gamma' : [0, 1] \to M'$ connecting $p$ and $q$ (which didn't pass through $B$) and then since connectedness and path-connectedness on manifolds are the same thing I'd have proven $M \setminus B$ was connected.
How can I rigorously show $M \setminus B$ is connected? If it's of any help $M \setminus B$ is a $n$-dimensional manifold with boundary homeomorphic to $\mathbb{S}^{n-1}$.
EDIT : Let $M$ be a $n$-manifold. A regular co-ordinate ball $B \subseteq M$ is a co-ordinate ball for which there exists a neighbourhood $B'$ of $\overline{B}$ and a homeomorphism $\phi: B' \to B_{(\mathbb{R}^n, d)}(x, r')$ that takes $B$ to $B_{(\mathbb{R}^n, d)}(x, r)$ and $\overline{B}$ to $\overline{B}_{(\mathbb{R}^n, d)}(x, r)$ for some $r' > r > 0$ and $x \in \mathbb{R}^n$
Basically a regular coordinate ball is a homeomorph $B$ of the standard coordinate ball contained in the interior of a larger such homeomorph $B'$. As $B'\setminus B$ is connected (as $n>1$) then if $M$ is connected then so is $M\setminus B$. Take a path in $M$ between two points of $M\setminus B$ and replace segments of it passing through $B$ by segments within $B'\setminus B$ (details omitted). Or else apply Mayer-Vietoris for reduced homology to $B'$ and $M\setminus \overline B$.