Proof that there are only finitely many $1$-cells that have a $0$-cell as a boundary point in a CW complex

Question

Below is a lemma from John Lee's Introduction to Smooth Manifolds.

Suppose $M$ is a $1$-manifold endowed with a regular CW decomposition. We want to show that every $0$-cell of $M$ is a boundary point of exactly two $1$-cells.

I cannot understand a line from the proof, namely:

It can be shown that the boundary of every $1$-cell of $M$ consists of exactly two $0$-cells. Now, suppose $v$ is a $0-$cell of $M$. Then why are there only finitely many $1$-cells that have $v$ as a boundary point?

Answer 1

If a 0-cell has no 1-cells attached, this point has no open set around it homeomorphic to $\mathbb{R}$ since it every small enough open set is a point. If a 0-cell has a single 1-cell attached, then every small enough open neighborhood looks like $[0,\infty)$ (since otherwise it would be attached on both ends by a 1-cell but this is not regular) which you can show is not homeomorphic to $\mathbb{R}$. If you have two or more 1-cells attached, you are a wedge of more than 2 copies of $[0,\infty)$ which can be shown to not be homeomorphic to $\mathbb{R}$.

Answer 2

Okay so after reviewing the chapter I figured it out. We need to use that the CW decomposition is locally finite. Proposition 5.16 of the text states that a CW complex is locally compact if and only if it is locally finite. A manifold is locally compact, so the CW complex here is locally finite. This means that $\{\bar{e}:e \in \mathscr{E}\}$, the collection of the closures of the cells is locally finite. Hence, $v$ has a neighborhood that intersects at most finitely many closures of cells, hence so does $v$.

Answer 3

Using algebraic topology there is another point of view to this problem, more abstract and general but very illuminating, and not requiring any extra hypothesis such as "local finiteness".

Let $X$ be a 1-dimensional CW complex, let $\{e_i\}_{i \in I}$ be the 1-cells, and for each $e_i$ let $\chi_i : [0,1] \to X$ be a characteristic map. For each $0$-cell $v$, define the valence of $v$ to be the cardinality of the set $\{(i,t) \in I \times \{0,1\} \mid \chi_i(t)=v\}$. Then one can prove that the valence of $v$ is equal to the rank of the free abelian group $$H_1(X,X-v) $$ In your problem, one can use any coordinate system on the 1-manifold $M$, combined with the excision theorem, to show that $H_1(M;M-v) \approx H_1(\mathbb R;\mathbb R-\{0\})$ which has rank $2$. The vertex $v$ therefore has valence equal to $2$. And as said, the boundary of a 1-cell of a regular CW decomposition of $M$ consists of two different $0$-cells, hence each $1$-cell has at most one of its two endpoints located at $v$. Therefore, there are exactly two 1-cells having endpoints at $v$.