Let $X$ be an $n$-dimensional CW-complex and let $A \subseteq X$ be a subcomplex. I want to show that the quotient space $X/A$ admits a structure of a CW-complex with skeletons $(X/A)^j := \pi(X^j \cup A)$, where $X^j$ is the $j$-skeleton of $X$ and $\pi : X \to X/A$ is the quotient map.
After some drawings, it became clear to me that the cells of $X/A$ are the cells of $X$ that do not meet $A$. I want to formalize this.
Let $e^j_1, \dots, e^j_{s_j}, \dots e^j_{r_j}$ be the $j$-cells of $X$, and suppose that $e^j_1, \dots, e^j_{s_j}$ do not belong to $A$, with $0 \leq s_j \leq r_j$. By definition, there is a map
$$f_j : (D_1^j \cup \cdots \cup D_{s_j}^j, S_1^{j-1} \cup \cdots \cup S_{s_j}^{j-1}) \to (X^j, X^{j-1})$$
that is a relative homeomorphism, where $D_i^j$ is a $j$-dimensional disk and $S_i^{j-1}$ is its boundary sphere (a continuous map of pairs $f:(A,B) \to (C,D)$ is a relative homeomorphism if $f : A \setminus B \to C \setminus D$ is a homeomorphism). My guess is now to define
$$g_j : (D_1^j \cup \cdots \cup D_{s_j}^j, S_1^{j-1} \cup \cdots \cup S_{s_j}^{j-1}) \to ((X/A)^j, (X/A)^{j-1})$$
as the composition of $f_j$ with the inclusion $(X^j, X^{j-1}) \hookrightarrow (X^j \cup A, X^{j-1} \cup A)$ and then with the projection $\pi$.
I am having trouble showing that $g_j$ is a relative homeomorphism, though.
Best Answer
If you trace through the definitions you will see that for each $i=1,...,s_j$, the restriction map $$g^j_i \mid D^j_i - S^{j-1}_i $$ is a continuous bijection onto its image. It is also an open map, because if the restriction of a quotient map to a saturated open subset is continuous and bijective then that restriction is an open map. Therefore it is a homeomorphism.