In the appendix to Algebraic Topology, Hatcher briefly goes over why
A subset $A$ of a CW complex $X$ is open in $X$ $\iff$ $\Phi_\alpha^{-1}(A)$ is open in $D_\alpha^n$ for all $\Phi_\alpha$, where $\Phi_\alpha:D_\alpha^n \to X$ is a characteristic map (the map which attaches a particular $n$-disk to $X$ continuously.)
The proof in the $(\implies)$ direction is quite easy, since the $\Phi_\alpha$ are continuous.
However, for the proof in the $(\impliedby)$ direction, I do not follow Hatcher completely. Quoting:
…and in the other direction, suppose $\Phi_\alpha^{-1}(A)$ is open in $D^n_\alpha$ for each $\Phi_\alpha$, and suppose by induction on $n$ that $A ~\cap~X^{n-1}$ is open in $X^{n-1}$. Then since $\Phi_\alpha^{-1}(A)$ is open in $D_\alpha^n$ for all $\alpha$, $A \cap X^n$ is open in $X^n$ by the definition of the quotient topology on $X^n$.
For reference, $X = \bigcup_n X^n$, and $X^n = (X^{n-1} \sqcup_{\alpha}D_\alpha^n)/(x \in \partial D_\alpha^n \sim \Phi_\alpha(x))$.
It is the bolded portion where I do not precisely follow. In more detail, what is Hatcher's reasoning? I believe what is going on is that because $\Phi_\alpha^{-1}(A\ \cap X^n)$ is open in every $D^n_\alpha$, and $\Phi_\alpha$ is just the quotient map $q:X^{n-1} \sqcup D_\alpha^n \to X^n$ composed appropriately with inclusions, we get$q^{-1}(A \cap X^n)$ is open. Is this approximately correct? Where can my understanding be made more precise?
Best Answer
A CW complex $X$ is endowed with the weak topology, meaning that, in order to show that $A$ is open in $X$, it suffices to show that $A \cap X^n$ is open in $X^n$. Now, the idea is to work your way up inductively. Suppose you know already that $A \cap X^{n-1}$ is open in $X^{n-1}$. How can we show $A \cap X^n$ is open in $X^n$?
We invoke the fact that $X^n$ is formed from $X^{n-1}$ as a quotient space $(X^{n-1} \sqcup_\alpha D^n_\alpha) / (x \in \partial D^n \sim \Phi_\alpha^n(X))$: it suffices to verify, by the quotient property, that $A$ is open in $X^{n-1}$ (but this we know by induction) and that its pre-images under $\Phi^n_\alpha$ are open in the $D_\alpha^n$. But this is our starting assumption.