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.
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).