Path component of a CW-complex is a subcomplex.

Question

Let $X$ be a CW-complex. I want to prove each of its path components contains a zero cell. For that, I want to prove that each of its components is again a CW-complex.

Please give me some hints.

Answer 1

Let C be your pathcomponent, and take any cell $e_n\to C$, which is, say, of dimension $n$. Then the boundary of $e_n$ maps to a cell of dimension ___?

Answer 2

A subcomplex of a CW-complex $X$ is a subspace $A \subset X$ which is a union of closed cells of $X$. Recall that a closed $n$-cell $e^n_\alpha$ is the image of an attaching map $\phi^n_\alpha : D^n \to X^{n-1}$ defined on the closed unit ball in $\mathbb{R}^n$. In particular, each $e^n_\alpha$ is pathwise connected. Note that if $e^n_\alpha \subset A$, then also all $e^m_\beta$ such that $\mathring{e}^m_\beta \cap e^n_\alpha \ne \emptyset$ must be contained in $A$. Here $\mathring{e}^m_\beta$ denotes the open cell associated to $e^m_\beta$, i.e. the set $\phi^m_\beta(\mathring{D}^m) \subset e^m_\beta$. $\mathring{D}^m$ is the open unit ball in $\mathbb{R}^n$.

Each path component $P$ of $X$ is a union of closed cells: If $e^n_\alpha \cap P \ne \emptyset$, then $e^n_\alpha \cup P$ is pathwise connected so that $e^n_\alpha \subset e^n_\alpha \cup P \subset P$.