Infinite CW Complex

Question

I recently picked up Hatcher's Algebraic Topology, and I'm struggling with the following notion in his definition of a CW complex in Chapter 0:

(3) One can either stop this inductive process at a finite stage, setting $X = X^n$ for some $n < \infty$, or one can continue indefinitely, setting $X = \bigcup_\alpha X^n$. In the latter case $X$ is given the weak topology: A set $A \subset X$ is open (or closed) iff $A \cap X^n$ is open (or closed) in $X^n$ for each $n$.

I'm having trouble getting a grip of the case where we don't stop the procedure at a finite final dimension. From a purely set-theoretical perspective, each $X^n$ is not contained in $X^m$ for $m > n$, since each $X^n$ is formally constructed as a quotient of $X^{n-1}$ and some $n$-discs. Therefore the union $\bigcup_n X^n$ is literally just a disjoint union of separate skeletons, which doesn't seem to be interesting or the point of the construction.

I feel like the real idea is to identify $X^n$ as a subspace of $X^{n-1}$ for all $n$ and then take the union. But how would we formalize this? It seems like there's going to be a lot of relabeling involved. I remember seeing a similar sort of construction in algebra that used the notion of a direct limit--is that the right way to look at this too?

Answer 1

Let $$\pi:X_{n-1}\sqcup\bigsqcup_{\mathcal{E}} D^n\rightarrow X_n$$ be the quotient map where $\mathcal{E}$ indexes the $n$-cells of $X$. Then we check that the restriction $\pi|_{X_{n-1}}$ is injective.

Let $A\subset X_{n-1}$. Then $\pi^{-1}(\pi(A))=A\sqcup\bigsqcup_{i\in\mathcal{E}}\varphi_i^{-1}(A)$, where $\varphi_i:S^{n-1}\rightarrow X_{n-1}$, $i\in\mathcal{E}$, is the attaching map of the $i^{th}$ cell. If $A$ is closed in $X$, then $\varphi_i^{-1}(A)$ is closed in $S^{n-1}$, and hence also in $D^n$. It folllows that if $A\subseteq X$ is closed, then $\pi^{-1}(\pi(A))$ is closed in $X_{n-1}\sqcup\bigsqcup_{\mathcal{E}} D^n$, and so $\pi(A)$ is closed in $X_n$, which is a quotient of the disjoint union. Thus:

The inclusion $X_{n-1}\hookrightarrow X_n$ is a closed embedding.

By induction we have that $X_n$ is closed in each $X_m$, $m\geq n$. If $m\leq n$, then $X_m\cap X_n=X_m$ is closed. Since $X=colim\;X_n$ carries the weak topology with respect to its skeleta we see that $X_n$ is embedded in $X$ as a closed subspace.

Answer 2

Starting with $X^n$ and a disjoint union of $n+1$ balls $B=\sqcup_{i\in I} B_i$, and maps $f_i\colon \partial B_i\to X^n$, we construct $X^{n+1}$ by quotienting out by the equivalence relation generated by $b\sim f(b)$ for all $b\in \partial B$.

The answer to your question is that two distinct points of $X_n$ are never related by this equivalence relation. A point $x\in X^n$ may be related to several $b\in B$ which map to $x$ under their respective boundary maps. The entire equivalence class of $x$ is then just $$\{x\}\cup \bigcup_{i\in I} f_i^{-1}(x).$$

Thus we get a sequence of nested inclusions $$X^0\subseteq X^1\subseteq X^2\cdots$$ and we may define $X$ as the union of these.