I am following Hatcher's book on algebraic topology. There I found the definition of subcomplex of a CW complex $X.$ The definition of subcomplex of a CW complex is given as follows $:$
Definition $:$ A subcomplex of a CW complex $X$ is a subspace $A \subseteq X$ which is a union of cells of $X,$ such that the closure of each cell in $A$ is contained in $A.$
From here the author claims that it is easy to see by induction over skeleta that a subcomplex is a closed subspace. But I don't get his point. Let $A$ be a subcomplex of a CW complex $X.$ Then in order to show that $A$ is closed in $X$ we need to show that $A \cap X^{n}$ is closed in $X^{n}$ for every $n \geq 0.$ It is clear that $A \cap X^{0}$ is closed in $X^{0},$ since $X^{0}$ is a discrete set. Now assuming that $A \cap X^{n-1}$ is closed in $X^{n-1}.$ We need to show that $A \cap X^n$ is closed in $X^n.$ Let $q : X^{n-1} \bigsqcup \left (\bigsqcup\limits_i D^n_i \right ) \longrightarrow X^n$ be the quotient map. So in order to show that $A \cap X^n$ is closed in $X^n$ it is enough to show that $q^{-1} (A \cap X^n)$ is closed in the coproduct (the domain of $q$). I think the pullback of $A \cap X^n$ lying in $X^{n-1}$ is $A \cap X^{n-1}$ which is closed in $X^{n-1},$ by induction hypothesis. Though I am not quite sure about that. Also what would be the pullback of $A \cap X^n$ lying in $\bigsqcup\limits_i D^n_i\ $?
Would anybody please help me in this regard? Thanks in advance.
EDIT $:$ Let $\{e^n_i\}_{i \in I_n}$ be the collection of open $n$-cells of $X$ and $\{e^n_i\}_{i \in J_n}$ be the collection of open $n$-cells of $X$ which are contained in $A.$ Consider the map $\Phi$ obtained by composing the inclusion map $\iota : \bigsqcup\limits_{i \in I_n} D^n_i \longrightarrow X^{n-1} \bigsqcup \left (\bigsqcup\limits_{i \in I_n} D^n_i \right )$ followed by the quotient map $q.$ Then the pullback of $A \cap X^n$ under $\Phi$ is $\Phi^{-1} (A \cap X^{n-1}) \bigsqcup \left (\bigsqcup\limits_{i \in J_n} \text {int} (D^n_i) \right ).$
Best Answer
Indeed, the component of $q^{-1}(A\cap X^n)$ in $X^{n-1}$ is $A\cap X^{n-1}$. This is essentially by definition, but let us spell it out, because Hatcher is not always explicit about this. In reality, $X^{n-1}$ is not a literal subset of $X^n$, instead we always canonically identify with a subset of $X^n$ by the composition $X^{n-1}\hookrightarrow X^{n-1}\sqcup\coprod_iD_i^n\twoheadrightarrow X^n$, where the first map is the inclusion and the second map is the quotient map and $A\cap X^{n-1}$ is precisely the preimage of $A$ (when considered a subset of $X^n$) under this map, i.e. it is the component of $q^{-1}(A\cap X^n)$ in $X^{n-1}$.
Now, fix an index $i$ and let $\Phi_i\colon D_i^n\rightarrow X$ be the characteristic map of $D_i^n$ and $e_i^n\subseteq X$ the corresponding $n$-cell. There are two options: 1.) $e_i^n\subseteq A$. Then $\overline{e_i^n}\subseteq A$ by definition of subcomplex, whence $\phi_i(D_i^n)=\phi_i(\overline{\mathrm{int}(D_i^n)})\subseteq\overline{\phi_i(D_i^n)}=\overline{e_i^n}\subseteq A$, so that the component of $q^{-1}(A\cap X^n)$ in $D_i^n$ is precisely $\phi_i^{-1}(A)=D_i^n$, so in particular closed. 2.) $e_i^n\cap A=\emptyset$ (this is a dichotomy, because we know that $A$ is a union of cells and the cells are pairwise disjoint). In this case, the component of $q^{-1}(A\cap X^n)$ in $D_i^n$ consists only of points on the boundary $\partial D_i^n$ that get mapped to $A$, but those are precisely the points in the preimage $\varphi_i^{-1}(A\cap X^{n-1})$, where $\varphi_i$ is the attaching map of $D_i^n$, so in particular closed. All in all, $q^{-1}(A\cap X^n)$ is closed in $X^{n-1}\sqcup\coprod_iD_i^n$ and the argument goes through.