My definition for a CW complex is any space built by the following inductive procedure:
- A $0$ dimensional CW Complex is a topological space with the discrete topology.
- Let $X_n$ be an $n$-dimensional CW complex and $\phi_n: \amalg S_n \to X_{n}$. The corresponding $n+1$ dimensional CW complex $X_{n+1}$ is given by the following pushout diagram:
\begin{align*} \require{AMScd} \begin{CD} \amalg S^n @>{}>> \amalg D^{n+1}\\ @V{\phi_n}VV @VV{}V \\ X_n @>>{}> X_{n+1} \end{CD} \end{align*}
An infinite dimensional CW complex $X$ can be formed as a colimit of finite ones. Given an inclusion $D^n \hookrightarrow \amalg D^n$ the characteristic map $\tau: D^n \to X$ is the map induced on $D^n$ s.t. the relavent diagram commutes. $\tau(D^n)$ is a cell and $\tau(int(D^n))$ is the corresponding open cell.
I get the definition and i have a solid intuition about CW complexes. I have a problem though whenever i try to prove something formal about them. I get confused by the diagrams. The question i'm stuck on now is:
How do I prove in the most simple and elegant way that open cells are disjoint?
I put some subscripts to the balls in order to distinguish them from each other. Here $\alpha$ uniquely determines the ball $D_\alpha^n$, so there is not really the need to keep the superscript $n$. This way, $\phi_n$ restricts on $S_\alpha^n$ to a gluing map $\phi_\alpha:S_\alpha\to X_n$. Let $r_\alpha$ denote the characteristic map for $D_\alpha$, and let $i_n:X_n\to X^{n+1}$ be the map induced by the pushout square.
\begin{align*} \require{AMScd} \begin{CD} \amalg S_\alpha^n @>{\amalg i_\alpha}>> \amalg D_\alpha^{n+1}\\ @V{\phi_n}VV @VV{r^{n+1}}V \\ X_n @>{i_n}>> X_{n+1} \end{CD} \end{align*}
It can be shown that since $\amalg i_\alpha$ is a (closed) inclusion, $i_n$ is a (closed) inclusion as well. This again imlies that the canonical map $j_n$ from $X_n$ to the colimit $X=\varinjlim X_n$ is a (closed) inclusion (cf. Directed limits of topological spaces and embeddings).
Now if $x$ is a point in the CW complex $X$, there is a minimal $m$ such that $x\in j_{m+1}(X_{m+1})$. Note that $X_{m+1}$ is a disjoint union of two sets: $i_m(X_m)$ and $r^{m+1}(\amalg \operatorname{int}(D_\alpha^{m+1}))$, and $$r^{m+1}(\amalg \operatorname{int}(D_\alpha^{m+1})) = \amalg r_\alpha(\text{int}(D_\alpha))$$ Since $x\notin j_m(X_m)$, it must lie in one open cell $j_{m+1}(r_\beta(\text{int}(D_\beta^{m+1})))$. If $j_p(r_\gamma(\text{int}(D_\gamma^p)))$ is a cell with $p>m+1$, then this cell is disjoint from $j_{p-1}(X_{p-1})$, which contains $j_{m+1}(X_{m+1})$ and thus $x$, so $j_{m+1}(r_\beta(\text{int}(D_\beta^{m+1})))$ is the unique cell which contains $x$.