The boundary of any manifold that is itself a boundary of a higher-dimensional manifold must always be empty?

Question

I have seen it asserted several times that it is well-known that the boundary of any manifold that is itself a boundary of a higher-dimensional manifold must always be empty. Yet, despite it supposedly being well-known, I have not been able to find any proof of this theorem. I have also searched through two popular topology textbooks – Topology by Munkres and Algebraic Topology by Hatcher – but have not been able to find this theorem. Is there any reputable source, such as a well-known topology textbook, that discusses this theorem and provides a proof? Or, if anyone here has a (relatively basic) proof of this theorem, I'd appreciate it.

Answer 1

This is the proposition on page 59, Chapter 2 in Guillemin and Pollack, Differential Topology:

If $X$ is a $k$-dimensional manifold with boundary, then $\partial X$ is a $(k-1)$ dimensional manifold without boundary.

A proof is given immediately after. Note the identification of manifolds without boundary and manifolds with empty boundary on the previous page.

The crucial point is that if $\varphi : U \to \mathbf{R}^{n-1} \times \mathbf{R}_{\ge 0}$ is a chart for $X$ (with $U \cap \partial X \ne \varnothing$) then the restriction $\varphi : U \cap \partial X \to \mathbf{R}^{n-1} \times \{0\}$ is a chart for $\partial X$.

Answer 2

@QiaochuYuan comment is very wise; too bad he had no time to expand it into a full answer.

My reference book here is "Topologie algébrique", Yves Félix & Daniel Tanré, in French.

(First: one must not confuse a boundary in general topology, and a boundary in algebraic topology. Although related, the two concepts have different definitions and properties. In general topology, in the usual case $\delta$ is idempotent: $\delta \circ \delta = \delta$; in algebraic topology, as you noted, $\delta \circ \delta = 0$. You are certainly aware of that, this is more a note for other readers).

The fact that $\delta \circ \delta = 0$ is key in homology. It is in fact part of the definition of a chain complex:
a sequence of $R$-modules$^{(*)}$ ($R$ is a commutative ring) $C_n$ and morphisms of $R$-modules $d_n$ :
$\dots \overset {d_{n+1}} \longrightarrow C_n \overset {d_n} \longrightarrow C_{n-1} \overset {d_{n-1}} \longrightarrow \dots \overset {d_1} \longrightarrow C_0 \overset {d_0} \longrightarrow 0$
where $\forall n \ge 1, d_n \circ d_{n-1} = 0$.
Boundary operator is $\delta = (d_n)_n$.

(*) This can also be presented with abelian groups, but $R$-modules are more general, and any abelian group can be made into a $\mathbb Z$-module by repeated addition, i.e. $n.x = x+x+ \dots +x$.

The $R$-module of $n$-degree cycles, $Z_n$, is by definition $d_n$ kernel.
The $R$-module of $n$-degree boundaries, $B_n$, is by definition $d_{n+1}$ image.
As $d_{n+1} \circ d_n = 0$, $B_n \subset Z_n$: all boundaries are cycles.
Quotient $H_n = Z_n / B_n$ is called the $n^{\text{th}}$ homology module of $(C_*,d)$ complex. If all cycles are boundaries, then $H_n=0$.
And then definitions go on with chain complex morphisms, short exact sequences of chain complexes, long exact sequences, etc. The fact that by definition $\delta \circ \delta = 0$ plays a key role in all results.

Now, how does that fit with geometric intuition?
In the case of simplicial homology, complexes are constructed using oriented simplexes.
If we denote by $\langle a_1, a_2, \dots, a_n \rangle$ the oriented simplex whose vertices are $a_1, a_2, \dots, a_n$,
for any permutation $\sigma$,
$\langle a_{\sigma(1)}, a_{\sigma(2)}, \dots, a_{\sigma(n)} \rangle = (-1)^{\varepsilon(\sigma)} \langle a_1, a_2, \dots, a_n \rangle$, where $\varepsilon(\sigma)$ is the sign of the permutation.

Then the boundary operator $\delta$ is defined by $\delta \langle a_1, a_2, \dots, a_n \rangle = \sum\limits_{k=1}^n (-1)^{k+1} \langle a_1, a_2, \dots, {{\overset - a}_k}, \dots, a_n \rangle$, where ${{\overset - a}_k}$ means: suppress $a_k$.
With this, when calculating $\delta \circ \delta$, terms cancel each other and we end with $\delta \circ \delta = 0$.
Intuitively, this comes from the fact that in an $n$-simplex, the $n-2$ simplexes are each present twice, as boundaries of two $n-1$-simplexes, and their orientation is opposite, so they cancel each other in $\delta \circ \delta$.

Singular homology shows similar definitions and results.

This intuition is of course less clear in the case of manifolds. However I am content with the fact that, in general, continuous theories can be seen as limits of discrete theories, and algebraic topology is full of such cases.