If manifolds $M_1$ and $M_2$ are connected, $M_1\# M_2$ is connected. Show that if $M_1$ and $M_2$ are connected $n$-manifolds and $n>1$, then $M_1 \# M_2$ is connected.
$M_1 \# M_2$ is the connected sum of the two manifolds.
This is problem 4.18(b) from Lee's Introduction to Topological Manifolds. My proof:
A theorem from the book furnishes, mutatis mutandis, topological embeddings $e_i:M_i^\prime\to M_1 \# M_2$ such that:
$$e_1(M_1^\prime) \bigcup e_2(M_2^\prime) = M_1 \# M_2$$
$$e_1(M_1^\prime) \bigcap e_2(M_2^\prime) = e_1(\partial M_1^\prime) = e_2(\partial M_2^\prime)$$
where $M_i^\prime = M_i \backslash B_i$ is the manifold with boundary that is formed by taking a regular coordinate ball from $M_i$ (the boundaries of the $M_i^\prime$ are joined to form the connected sum).
Since the $M_i^\prime$ have nonempty boundary, $e_1(M_1^\prime) \bigcap e_2(M_2^\prime) \neq \emptyset$. Now, $e_i(M_i^\prime)$ being a topological embedding, it is connected if and only if $M_i^\prime$ is connected, which is true if $M_i$ is connected.
So assuming that $M_1$ and $M_2$ are connected, this shows that $M_1 \# M_2$ is the union of nonempty connected sets with nonempty intersection, which implies that it is connected.
This seems to be correct but I never had to assume that $n>1$, which troubles me. Is this condition necessary? Intuitively, it doesn't seem to be, so I'm confused. Let me know where I went wrong, if I did, but please don't just give the correct answer.
 A: Take the following manifolds $(0,3)\times \{0\}$, $(0,3)\times \{1\}$. Remove the balls $(1,2)\times \{0\}$ and $(1,2)\times \{ 1\}$. Now glue $(0,1)\times\{0\}$ to $(0,1)\times \{1\}$ and $(2,3)\times\{0\}$ to $(2,3)\times \{1\}$. Then you end up with two lines, which is not connected.
Added:
The problem is (as Pedro already said in the comment) that in dimension one removing a ball may disconnect the manifold (which is exactly what happens in the example above).
A: Your reasoning is correct. The dimensional restriction comes in when you arguing that $M_i\smallsetminus B_i$ is connected when $M_i$ is. To show this, it is equal to show that $M_i \smallsetminus B_i$ is path-connected. When we're done showing this, the rest of the arguments follows the same route as yours.
To show $M \smallsetminus B$ is path-connected, 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 $\mathbb{S}^{n-1}$, and hence $\partial \bar{B}$ is connected (this is where the dimensional restriction comes in). 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 $q$.
