Every characteristic function $\Phi_{\beta}^b: E^n_{\beta} \to e^{-n}_{\beta}$ is an identification function.
My book says the following:
This follows from the fact the the CW complex $X$ has $X^n$ a quotient space of $X^{n-1} \sqcup \coprod_{\beta}E^n_{\beta}$.
I am not sure I see how this follows. We know that $\Phi_{\beta} |S^{n-1}_{\beta} = \phi_{\beta}:S^{n-1}_{\beta} \to X^{n-1}$, but how do I use what the book gave me to prove it?
Note: $e^{n}_{\beta}$ is an open $n$-cell, $E$ is the unit ball, and $S$ is the unit sphere. The characteristic function is what I have defined above.
I don't see an easy way to prove this using what the books says. But if you know that CW complexes are Hausdorff, you can instead prove it as follows. Let $Y$ be the quotient space of $E^n_\beta$ by the equivalence relation induced by $\Phi^b_\beta$, i.e. $x\sim y$ iff $\Phi^b_\beta(x)=\Phi^b_\beta(y)$. Then $Y$ is compact since $E^n_\beta$ is, and $\Phi^b_\beta$ gives a continuous bijection $f:Y\to \overline{e^n_\beta}$. Since $X$ is Hausdorff, so is $\overline{e^n_\beta}$, and so $f$ is automatically a homeomorphism (it is a continuous bijection from a compact space to a Hausdorff space). But $\Phi^b_\beta$ is just $f$ composed with the quotient map $E^n_\beta\to Y$, so this means $\Phi^b_\beta$ is also a quotient map.