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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 $x \in \mathbb{R}^n$

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    $\begingroup$ What actually is a "regular" coordinate ball? I can see an interpretation of this term under which the conclusion is false. $\endgroup$ – Lord Shark the Unknown Dec 14 '17 at 5:50
  • $\begingroup$ @LordSharktheUnknown I've added the definition above $\endgroup$ – Perturbative Dec 14 '17 at 5:54
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    $\begingroup$ Your idea seems good to me: take a path from $p$ to $q$, and if it goes through $B$ then replace part of it with a path in the annulus $B' \setminus B.$ $\endgroup$ – Anthony Carapetis Dec 14 '17 at 6:05
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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 $M\setminus \overline B$.

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Your idea is good. Just for completeness i'll post an answer. Let $p,q \in M'=M\smallsetminus B$ be arbitrary and let $\alpha : [0,1] \to M$ be a path connecting them (guaranteed by hypothesis). If this path does not intersect $B$, then we are done. That is we obtain a path in $M'$ connecting $p$ and $q$. If it does intersect $B$, we do the following: let $t_1$ and $t_2$ be the minimum and the maximum of the closed subset $\alpha^{-1}(\bar{B}) \subseteq [0,1]$ respectively. That is $\alpha(t_1)$ is the point where the path hit $\bar{B}$ for the first time and $\alpha(t_2)$ is the point where the path is on $\bar{B}$ for the last time. Both of this points must lie on the boundary $\partial \bar{B}$. Since $B$ is a regular coordinate ball, $\partial \bar{B}$ is homeomorphic to $\mathbb{S}^{n-1}$ (this is where the dimensional restriction comes in), and hence $\partial \bar{B}$ is connected. Choose a path (after rescaling the parameter) $\beta : [t_1,t_2] \to \partial \bar{B}$ connecting $\alpha(t_1)$ and $\alpha(t_2)$, define a new path $\alpha' : [0,1] \to M \smallsetminus B$ as \begin{equation*} \alpha'(t) = \left\{ \begin{array}{rl} \alpha(t) & \text{for } 0 \leq t \leq t_1,\\ \beta(t) & \text{for } t_1 \leq t \leq t_2,\\ \alpha(t) & \text{for } t_2 \leq t \leq 1. \end{array} \right. \end{equation*} Therefore we obtained a path in $M'$ joining $p$ and $q$.

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