I am trying to prove (in a strict way) that $X/A$ inherits a natural CW-complex structure from a CW-pair $(X,A)$.
I start with the following inductive definition of a CW-complex:
Let $\{\mathcal A_n\}_{n=0}^\infty$ be a sequence of disjoint sets such that $\mathcal A_0\ne\emptyset$. Starting with this sequence we inductively construct a sequence of topological spaces $\{X^n\}$ as follow:
- For $n=0$ we put $X^0=\mathcal A_0$ with the discrete topology on $\mathcal A_0$.
- If $X^{n-1}$ has already been constructed, then put $X^n=X^{n-1}$ if $\mathcal A_n=\emptyset$. However if $\mathcal A_n\ne\emptyset$, we assume that we have a family of maps $\{\varphi_\alpha^n:\partial D_\alpha^n\to X^{n-1}:\alpha\in\mathcal A_n\}$ and define: $$X^n=\Big(X^{n-1}\bigsqcup_{\alpha\in\mathcal A_n}D_\alpha^n\Big)/\sim_n$$ where $\sim_n$ is the equivalence relation generated by $x\sim \varphi_\alpha^n(x)$ if $x\in \partial D_\alpha^n$.
Then a CW-complex $X$ is the colimit of the sequence of topological embeddings $X^0\hookrightarrow X^1\hookrightarrow\ldots$ as defined above
In order to prove that $X/A$ is a CW complex I want to show the following:
The quotient space $\tilde X^n=X^n/A$ determines a sequence of topological ${X^n}$ as the definition above.
I think that the best way to do this is proceed inductively
- First show that $\tilde X^0=X^0/A$ is a discrete space (trivial) and that $\tilde X^1=X^1/A$ is homeomorphic to $\big(\tilde X^0\bigsqcup_{\alpha\in\mathcal A_0}D_\alpha^n\big)/\sim_0$
- Suppose $\tilde X^n=X^n/A$ is homeomorphic to $\big(\tilde X^{n-1}\bigsqcup_{\alpha\in\mathcal A_n}D_\alpha^n\big)/\sim_n$ and conclude $\tilde X^{n+1}$ is homeomorphic to $\big(\tilde X^n\bigsqcup_{\alpha\in A_{n+1}}D_\alpha^{n+1}\big)/\sim_{n+1}$
Any help? How can I construct these homeomorphisms?