Suppose $Y$ is a $CW$ complex, of dimension $p-1,\ \sum B_{\alpha}$ is a topological sum of closed $p-$ balls. Then, if $g:\sum \partial B_{\alpha}\to Y$ is a continuous map, the adjunction space $X=Y\bigcup_g \sum B_{\alpha}$ is a $CW$ complex and $Y$ is its $p-1$ skeleton.
The proof seems straightforward enough, but I am stuck on several points. Let $f:Y\bigsqcup \sum B_{\alpha}\to X$ be the quotient map. $Y$ is identified with a subspace of $X$ since $f\restriction_ Y$ is a homeomorphism. Then, $f$ is inclusion on $Y$ and
now Munkres says $f=g$ on $\sum B_{\alpha}$ but since $x\mapsto [x]=\left \{ g(x) \right \}\cup \left \{ g^{-1}(g(x)) \right \}$ if $x\in \sum B_{\alpha}$, how can this be true?
From here, the cells in $X$ are defined in the obvious way: take the cells of $Y$ (of dimension $<p$), and $e_{\alpha}=f(B^{\circ}_{\alpha}).$ Now, $B^{\circ}_{\alpha}$ is clearly open in $Y\bigsqcup \sum B_{\alpha}$ but I do not see why it is saturated with relative to $f$, (and this is necessary to complete the proof, which is more or less a standard check that the definitions of a $CW$ complex are satisfied.)
edit: I think I see it now: $f$ maps each element of the boundary of $B_{\alpha}$ to its equivalence class, while $f(\text{int} B_{\alpha})$ is a set of singletons in $X$ and so int$B_{\alpha}$ is saturated with respect to $f$.
Is this correct?
Best Answer
Yes, it is correct. For the sake of simplicity let us consider the attachment of a single $p$-cell to the $(p-1)$-skeleton $Y$. In this case a closed ball $D^n \subset \mathbb{R}^n$ is attached to $Y$ via a map $\phi : S^{n-1} \to Y$. This yields the quotient space $Y \cup_\phi D^n = Y \sqcup D^n/\sim$, where $x \sim \phi(x)$ for $x \in S^{n-1}$. The quotient map $\pi : Y \sqcup D^n \to Y \cup_\phi D^n$ identifies points of the boundary $S^{n-1} \subset D^n$ in a nontrivial way with points of $Y$ and maps the interior $\mathring{D}^n$ of $D^n$ homeomorphically onto $\phi(\mathring{D}^n) \subset Y \cup_\phi D^n$. However, it is not excluded that $\phi$ is embedding in which case $p$ embeds $D^n$ into $Y \cup_\phi D^n$.