The inductive construction of a CW complex. I am confused about the following proposition from Lee's introduction to topological manifolds.

Proposition: Lex X be a CW complex. Each skeleton $X_n$ is obtained from $X_{n-1}$ by attaching a collection of $n$-cells.

The proof is in Introduction to Topological Manifolds page 138, proposition 5.18.
It claims that it suffices to show that the map $\Phi : X_{n-1} \sqcup (\bigsqcup_{\alpha}D^n_{\alpha}) \to X_n$ is a quotient map, where $D_\alpha^n$ are the $n$-cells. Why it suffices to do so?
I think there is a map from $X_{n-1} \bigcup_\varphi (\bigsqcup_{\alpha}D^n_{\alpha})$ to the space   $X_n$ by the universal mapping property. And my approach is trying to show that this map is a homeomorphism.
 A: (I update this post as soon as I found out how to draw good diagrams here.)
This is more or less an exercise with univeral properties. For a topological spaces $M,N$ and a quotient map $\pi:M\to N$ the following is true. Given a continous map $f:M\to Z$ with the property $f(\pi^{-1}(n))=z$ (i.e. $f$ is constant on preimages of the quotient map), then there exists a unique map $\tilde{ f}:N\to Z$ s.t. $\tilde{f}\circ \pi= f$.
Now suppose there is another quotient map $\pi':M\to N'$ s.t. said property is pairwise fullfilled. Then you have that there are unique maps $\tilde{\pi}:N\to N'$ and $\tilde{\pi'}:N'\to N$ commuting with the projections $\pi,\pi'$.
Now $\tilde{\pi'}\circ\tilde{\pi}:N\to N$ is an unique map commuting with $\pi$ but there is another map from $N\to N$ satisfying this namely the identity. Therefore $\tilde{\pi'}\circ\tilde{\pi}=id_N$ and similarly $\tilde{\pi}\circ\tilde{\pi'}=id_{N'}$. Therefore $\tilde{\pi}$ is an isomorphism. 
Now the claim in the book is that $\Phi : X_{n-1} \sqcup (\bigsqcup_{\alpha}D^n_{\alpha}) \to X_n$ and $\varphi : X_{n-1} \sqcup (\bigsqcup_{\alpha}D^n_{\alpha}) \to X_{n-1} \bigcup_\varphi (\bigsqcup_{\alpha}D^n_{\alpha})$ are quotient maps which satisfy: $$\varphi(x)=\varphi(y) \Leftrightarrow \Phi(x)=\Phi(y).$$
This brings you in the above situation and therefore $X_n\cong X_{n-1} \bigcup_\varphi (\bigsqcup_{\alpha}D^n_{\alpha})$.
