Why is the weak topology defined on cell complexes equivalent to the topology defined by the quotient map? Just for background, I have been reading from Hatcher's textbook where he defines a cell complex in the following way:

(1) Take any discrete set $X^0;$ we will call each element in this set a $0\textrm{ -}$cell.


(2) Let $\varphi_\alpha:S^{n-1}\rightarrow X^{n-1}$ be some collection of maps. Construct a quotient space $X^n$ of the space $X^{n-1}\bigsqcup_\alpha \mathbb{D}_\alpha^n$ by the relation $\sim$ defined as $x\sim \varphi_\alpha(x)$ for each $x\in\partial\mathbb{D}_\alpha^n.$ The space $X^n$ is called the $n\textrm{ -}$skeleton, and this process is described as attaching $n\textrm{ -}$cells $e_\alpha^n$ through the maps $\varphi_\alpha.$ Hence, we can write $X^n=X^{n-1}\bigsqcup_\alpha e_\alpha^n,$ where $\{e_\alpha^n\}$ is some collection of open $n\textrm{ -}$cells.


(3) Finish the construction by stopping the inductive process for some $n$ and let $X=X^n$ or let process go on infinitely and set $X=\bigcup_{n\in\mathbb{N}} X^n;$ we give the space the weak topology defined as $$\mathcal{T}=\{A\subseteq X:A\cap X^n\textrm{ is open in } X^n,\textrm{ for all }n\}.$$

I have questions on the topology defined above in a finite case, that is, when $X=X^n;$ for the quotient map  $q:Y\rightarrow Y/\sim,$ from any topological space $Y$ into $Y/\sim,$ the quotient space,  we define the quotient topology to be the collection of sets $\mathcal{U}=\{A\subseteq Y/\sim:q^{-1}(A) \textrm{ is open in } Y\}.$ I don't see how these two definitions, that is the (3) and the quotient topology are equivalent; we would need to prove that for $A\subseteq X$ and each $0\leq i\leq n,$ $$q^{-1}(A) \textrm{ is open in } X^{n-1}\bigsqcup_\alpha\mathbb{D}_\alpha^n \Longleftrightarrow X^{n-i}\cap A \textrm{ is open in } X^{n-i}.$$
Now, if $q^{-1}(A)$ is open in the original space, the implication easily follows; I don't see it forming in the opposite direction, why does it follow that if $X^{n-i}\cap A$ is open in $X^{n-i}$ where $0\leq i\leq n$ imply that $q^{-1}(A)$ is open in the original space?
 A: Hatcher describes how CW-complexes are constructed:

*

*Start with a set $X^0$ having the discrete topology.


*Construct inductively skeleta $X^n$ by attaching $n$-cells to $X^{n-1}$. Here $X^{n-1}$ already has a topology and $X^n$ is defined as a suitable quotient space of $X^{n-1}\bigsqcup_\alpha \mathbb{D}_\alpha^n$.
This construction produces a sequence of spaces $X^0, X^1, X^2,\ldots$ such that $X^{n-1}$ is a subspace of $X^n$.
If this process stops at some finite $N$, then we have a topology on $X = X^N$. In that case trivially $A  \subset X$ is open iff $A \cap X^n$ is open in $X^n$ for all $n$. Just note that $X^n = X$ for $n \ge N$.
If the attaching process continues ad infinitum, then we have a topology on each skeleton $X^n$, and these toplogies are compatible in the sense that $X^{n-1}$ is a subspace of $X^n$ for each $n$. Howewer, we do not have a topology on $X = \bigcup_{n=0}^\infty X^n$. That is why we give the space $X$ the weak topology defined as in (3). This definition is relevant only in case that the attaching process continues ad infinitum.
Edited on request:
Suppose we have an ascending sequence of topological spaces $X^0, X^1, X^2,\ldots$ (ascending means that for all $n > 0$ we have $X^{n-1} \subset X^n$ and that the topology on $X^{n-1}$ is the subspace topology inherited from $X^n$). Then let $X = \bigcup_{n=0}^\infty X^n$. Which topology do we introduce on $X$ in order that all subspaces $X^n$ receive their original topology? In general there are many ways to do that, but the standard approach is to define $A \subset X$ open in $X$ iff $A \cap X^n$ is open in $X^n$ for all $n$. This is the final topology with respect to the system $\{X^n\}$. In the context of CW-complexes it is also denoted as the weak topology. This has historical reasons. See Confusion about topology on CW complex: weak or final?
Let us prove that if $X$ is endowed with the final topology, then all subspaces $X^n \subset X$ have their original toplogy:

*

*Let $U \subset X^n$ be open in the subspace toplogy. Then $U = A \cap X^n$ with some open $A \subset X$. But by definition of the final topology $A \cap X^n$ is open in $X^n$ with its original toplogy.


*Let $U = U_n \subset X^n$ be open in the original topology. Using the fact that each $X^{k-1}$ is a subspace of $X^k$, we can recursively construct open $U_k \subset X^k$, $k \ge n$, such that $U_{k+1} \cap X^k = U_k$. Moreover, for $k < n$ define $U_k = U_n \cap X^k$ which is open in $X^k$. By construction we have $U_k \cap X^m = U_k$ if $k \le m$ and $U_k \cap X^m = U_m$ if $k > m$. Then $A = \bigcup_{k=0}^\infty U_k$ is open in $X$: In fact, $A \cap X_m = (\bigcup_{k=0}^\infty U_k) \cap X_m = \bigcup_{k=0}^\infty U_k \cap X_m = \bigcup_{k=0}^m U_k \cup \bigcup_{k=m+1}^\infty U_m = U_m$ which is open in $X^m$. This also shows $A \cap X^n = U_n = U$, i.e $U$ is open in the subspace topology.
If the sequence $X^0, X^1, X^2,\ldots$ stabilizes, i.e. if we have $X = X^N$ for some $N$ (which is the same as $X^n = X^N$ for $n \ge N$), then there is no need to introduce a new topology on $X$ since $X^N$ already has one. The above proof shows nevertheless that $A \subset X$ is open iff $A \cap X^n$ is open in $X^n$ for all $n$. However, in that case it is trivial: If $A \cap X^n$ is open in $X^n$ for all, then $A = A \cap X = A \cap X^N$ is open in $X = X^N$. Conversely, if $A \subset X^N$ is open, then $A \cap X^n$ is open in $X^n$ for all $n \le N$ because $X^n$ is a subspace of $X^N$ and open in $X^n$ for all $n > N$ because $X^n = X^N$ and $A \cap X^n = A$.
