"Any closed, oriented $3$-manifold $M$ is the boundary of some oriented $4$-manifold $B$." See this post: Elegant proof that any closed, oriented 3-manifold is the boundary of some oriented 4-manifold?
I heard this statement is true:
- (1) Any closed 3-manifold is a boundary of some compact 4-manifold.
- (2) Any 3-manifold $M$ can be realized as the boundary of a 4-manifold $B$.
In particular, we know that all 3-manifolds can be triangulable. However for 4-manifolds, there are simply connected non-triangulable manifolds (such as the E$_8$ manifold). (Note: a closed 4-manifold is triangulable if and only if it's smoothable.) See this MO post: Not all manifolds can be triangulated
(3) For any 3-manifold $M_3$ that can be realized as the boundary of a 4-manifold $B_4$, the $M_3$ must be triangulable. So must the $M_3$ be the boundary of a triangulable 4-manifold $B_4$?
(4) Are there any non-triangulable 4-manifold $B_4'$ with a 3-dimensional boundary (i.e. $B_4'$ is not closed)? Then would the 3-manifold boundary $M_3'$ be triangulable (if $M_3'$ is non-triangulable, isn't that leads to a contradiction)?
Can one show these (1), (2) and explain them as intuitively as possible?