boundary of a simply connected, compact manifold with boundary For a simply connected, compact n-manifold with boundary (n > 1), is its boundary connected?
It’s obviously false when n = 1, but how to prove or disprove the statement when n > 1? I’m especially interested in n = 2. Thanks.

$\require{AMScd}$
I believe that for $n=2$ there are no such manifolds $M$, at least if you require $M$ to be orientable. Assume such an $M$ exists and assume that we can decompose the boundary as $\partial M = U \sqcup V$, where now $U,V$ are connected $1$-manifolds. Then, using Poincaré-Lefschetz (which requires the orientability), we have the following commutative diagram, where the rows are long exact sequences of the pair $(M,\partial M)$:
\begin{CD}
\cdots @>>> H^1(M) @>>> H^1(\partial M) @>>> H^2(M,\partial M) @>>> \cdots\\
& @V\cong VV  @V\cong VV  @V\cong VV \\
\cdots @>>> H_1(M,\partial M) @>>> H_0(\partial M) @>>> H_0(M) @>>> \cdots\\
\end{CD}
Now we observe that:

*

*$H^1(M) \cong 0$, since $M$ is simply connected.

*$H_0(M) \cong \mathbb{Z}$, since $M$ is in particular connected.

*$H_0(\partial M) \cong \mathbb{Z}^2$, by definition of $\partial M$.

It follows that $H_1(M, \partial M) \cong 0$ and hence by exactness the homomorphism
$$
H_0(\partial M) \cong \mathbb{Z}^2 \to \mathbb{Z} \cong H_0(M)
$$
is injective. But this is a contradiction, because there are no injective homomorphisms $\mathbb{Z}^2 \to \mathbb{Z}$.
I hope this was helpful!
Edit: As user ecrin points out, the orientability of $M$ follows from it being simply connected. Then my proof should work in general.
