CW -complex structure of boundary of a manifold

Question

Given a CW-complex structure of manifold with boundary. Is there any natural way to construct CW-complex structure of its boundary?

Thanking you.

Answer 1

Just take the subcomplex of stuff lying within the boundary?

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As suggested in the comments, this begs the question of whether the boundary must actually be a subcomplex.

Lemma 1. Let $x\in M$. Then there are only finitely many cells $C$ with $x\in\partial C$.

Proof. From the atlas of $M$ pick a map $\phi$ around $x$ that maps $x\mapsto 0$. From each $C$ with $x\in \partial C$, pick an interior point $x_C$ with $|\phi(x_C)|<1$. By the definition of CW complex, the set of these $x_C$ is closed, but the set of all $\phi(x_C)$ is closed in $\mathbb R^n$ only if it is finite. $_\square$

Lemma 2. Let $A\subseteq M$ be subset homeomorphic to (an open subset of) $\mathbb R^k$. Then the intersection of $A$ with all $\ge k$-dimensional cells is dense in $A$.

Proof. For any point in $A$ (or via homeomorphism equivalently: point in $\mathbb R^k$), the (by lemma 1) finitely many $<k$-dimensional cells touching that point cannot cover the $k$-dimensional $A$. $_\square$

Lemma 3. Let $C$ be a cell that intersects $\partial M$. Then $\dim C\le n-1$.

Proof. If $a\in\partial M$, the no neighbourhood of $a$ is homeomorphic to an open subset of $\mathbb R^n$. $_\square$

Lemma 4. Let $C$ be a cell with $\dim C=n-1$ that intersects $\partial M$. Then $C\subseteq \partial M$.

Proof. Assume otherwise. Then $\partial M\cap C$ is a proper nonempty closed subset of $C$, hence has nonempty boundary ($C$ is connected!). Let $a$ be a point in this boundary. Via a map from the atlas of $M$, we may view a portion of $\partial M$ around $a$ as hyperplane in $\mathbb R^n$. Subtracting $C$ from this hyperplane, we are left with an open subset of $\mathbb R^{n-1}$ that has $a$ in its boundary. By lemma 2, there is a $\ge(n-1)$-dimensional cell $C'\ne C$ with $a\in\partial C'$. By lemma 3, $\dim C'=n-1$. But then by the definition of CW complex, $a\in\partial C'$ implies that the unique cell $C$ with $a\in C$ must have dimension $<n-1$, contradicton $_\square$

Now we note that lemma 2 implies (locall, i.e., map-wise) that the $(n-1)$-dimensional cells intersecting $\partial M$, together with the smaller cells making up their boundaries, cover $\partial M$. By lemma 4, these $(n-1)$-cells (and by closedness of $\partial M$ also the smaller cells making up their boundaries) are contained in $\partial M$. We conclude that these cells make up a CW complex of $\partial M$.

Answer 2

CW complexes are often constructed by defining their skeleta inductively. We can begin by taking the 0-skeleton to be a discrete space and attach the 1-cells to the 0-skeleton. So, each 1-cell would begin as a closed 1-ball and is attached to the 0-skeleton through a map, usually continuous, from the 0-sphere $S_0$.

Each point of the 0-sphere $S_0$ can be identified through its image in the 0-skeleton, which is a key example of an equivalence relation. The 1-skeleton would then be defined to be an identification space derived from the union of the 0-skeleton and their respective 1-cells by using this equivalence relation.