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I have a hard time understanding what the topology on CW-complexes is, i.e. what the open sets of a given CW-Complex $K$ or $n$-skeleton $K^{(n)}$ are.

From my understanding we get the topology inductively and we start with the discrete topology on the $0$-cells $K^{(0)}$.

Then we add the $1$-cells by taking the disjoint union with the $0$-cells and identifying the boundary of our $1$-cells with $0$-cells via the maps $f_\alpha: \partial D^1_{\alpha}\rightarrow K^{(0)}$.

And on $K^{(1)}$ we have the quotient topology. So this works for $n=1,2$ maybe $3$ but for general $n$ is there no easier way to determine whether a set is open in $K^{(n)}$ or not? An easier equivalent definition? Or is this actually easy already and i just don't really understand it? I guess my question is just this: what does it mean if $A \subset K^{(n)}$ is open?

Ton910
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1 Answers1

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There is a non-inductive description of $K^{(n)}$ which gives a direct description of the open sets.

Let the open $i$-cells be indexed as $\{e_{i,j} \mid j \in A_i\}$, with pairwise disjoint index sets $A_i$ for $i \ge 0$. Choose a characteristic map corresponding to each $e_{i,j}$ denoted $\chi_{i,j} : D^i \to K^{(i)}$. Form the disjoint union of all the domains of all of these characteristic maps: $$\Delta = \coprod_{0 \le i \le n} \coprod_{j \in A_i} D^i_{i,j} $$ where $D^i_{i,j}$ denotes (a copy of) the domain $i$-disc for the characteristic map $\chi_{i,j}$. Define $\chi : \Delta \to K^{(n)}$ to be the disjoint union of the characteristic maps $\chi_{i,j}$ themselves. Then the topology on $K^{(n)}$ is just the quotient topology induced by $\chi$, i.e. it is the weakest topology on $K^{(n)}$ such that $\chi$ is continuous (equivalently such that each $\chi_{i,j}$ is continuous). Thus, a subset of $K^{(n)}$ is open if and only if its inverse image under $\chi$ is open.

I believe you can find this point of view explained in an appendix of Hatcher's "Algebraic Topology".

And this point of view actually works for the whole CW complex even in the infinite dimensional case, you simply use $$\Delta = \coprod_{0 \le i < \infty} \coprod_{j \in A_i} D^i_{i,j} $$

Lee Mosher
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  • Thanks a lot, this is more intuitive to me. So if i'm understanding correctly this means that: A subset $A \subset K^{(n)}$ is open if and only if for every cell $e_{i,j}$ the preimage $\chi_{i,j}^{-1}(A)$ is open. Is that correct? – Ton910 Jan 23 '21 at 14:22
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    That's correct. – Lee Mosher Jan 23 '21 at 15:25