A confusion in showing that the map $\Phi: X_{n-1} \amalg (\amalg_\alpha D_\alpha^n)\to X_n$ is quotient Let$X$ be a CW complex and $\{e_\alpha^n\}$ be the collection of $n$-cells of $X$ and for each $n$-cell $e_\alpha^n$ let $\Phi_\alpha^n: D_\alpha^n \to X$ be a characteristic map. Define the map $\Phi: X_{n-1} \amalg (\amalg_\alpha D_\alpha^n)\to X_n$ that is equal to inclusion on $X_{n-1}$ and to $\Phi_\alpha^n$ on each $D_\alpha^n$.
Suppose that $A$ is a saturated closed subset of the disjoint union, and let $B=\Phi(A)$, so that $A = \Phi^{-1}(B)$. The hypothesis means that $A \cap X_{n-1}$ is closed in $X_{n-1}$ and $A \cap D_\alpha^n$ is closed in $D_\alpha^n$ for each $\alpha$.
In this case, how does the first assertion imply that $B \cap \bar{e}$ is closed in $\bar{e}$ for every cell $e$ of dimension less than $n$; and the second imply that $B \cap \bar{e_\alpha^n}$ is closed in $\bar{e_\alpha^n}$ for each $n$-cell because $\Phi_\alpha^n: D_\alpha^n \to \bar{e_\alpha^n}$ is a closed map? 
I think they use something like $\Phi(A \cap X_{n-1} \cap \bar{e}) = \Phi(A \cap \bar{e}) = \Phi(A) \cap \Phi(\bar{e}) = B \cap \bar{e}$ but we can split the intersection from the function only if the function is injective. In the first case, maybe we can get this because $\Phi$ restricted to $X_{n-1}$ is inclusion, but in the second case $\Phi_\alpha^n$ is not injective on $A \cap \partial D_\alpha^n$, so how do we get this result?
 A: You are right about the first part. For the second, note that
$\Phi(A\cap D_{\alpha}^n)=\Phi(\Phi^{-1}(B)\cap D_{\alpha}^n))$ is closed in $\overline e_{\alpha}^n$ so it suffices to prove that $\Phi(\Phi^{-1}(B)\cap D_{\alpha}^n))=B\cap \Phi(D_{\alpha}^n)$ since $\Phi(D_{\alpha}^n)=\overline e_{\alpha}^n.$ 
So, for arbitrary $x\in B\cap \Phi(D_{\alpha}^n)$, we need to find $a\in \Phi^{-1}(B)\cap D_{\alpha}^n$ such that $\Phi(a)=x.$ Now, there is $a'\in A$ such that $x=\Phi(a')$. If $a'\in D_{\alpha}^n,$ we are done. And if $a'\notin D_{\alpha}^n$, then $x=a'=\Phi(a')=\Phi(d)$ for some $d\in  D_{\alpha}^n$ and again we are done. $\left(a'\in \Phi^{-1}(B)\Rightarrow \Phi(a')\in B\Rightarrow \Phi(d)\in B\Rightarrow d\in \Phi^{-1}(B)\right).$
The other inclusion is immediate. 
A: *

*If $e$ is a cell of dimension less than $n$, then it is contained in $X_{n-1}$. Since $X_{n-1}$ is closed in $X$, the closure $\bar{e}$ also must be contained in $X_{n-1}$. Therefore, both $A\cap \bar{e}$ and $B \cap \bar {e}$ are contained in $X_{n-1}$, and the former is a closed set in $\bar{e}$, because it equals the union of $\bar{e}$ with the closed set $A\cap X_{n-1}$. Now I believe you can show the sets $A \cap \bar{e}$ and $B \cap \bar{e}$ are equal (Use the fact that $\Phi|_{X_{n-1}}$ is the identity).

*For the second one, since the map $\Phi ^n _\alpha :D_\alpha ^n \to \bar{e}^n _\alpha$ is a closed map, it suffices to show that the image of the closed subset $A\cap D^n_\alpha \subset D^n_\alpha$ under $\Phi^n_\alpha$ equals $B \cap \bar{e}^n_\alpha$, i.e., $\Phi^n_\alpha(A\cap D^n_\alpha)=B \cap \bar{e}^n_\alpha$. But, this is easy, just use the facts that $B=\Phi(A)$ and that $A=\Phi^{-1}(B)$.
I'm sure that you are reading the book "Introduction to Topological Manifolds". It is a good book (I personally think chapters 3-5 are the best).
