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I want to prove the following little Lemma:

Let K be a CW-complex and let $S \subset K $. Prove that: S is open in K if and only if $f_{\sigma}^{-1}(S)$ is open for each $\sigma$. Where: $f_{\sigma}:B_{\sigma}^n \to K$ is the characteristic map of the cell $\sigma$.

I know that the characteristic map is continuous. Therefore if S is open, $f_{\sigma}^{-1}(S)$ is open. I found the following proof of the other direction: Suppose that $ f_{\sigma}^{-1}(S)$ is open in $B_{\sigma}^n$ for each $f_{\sigma}$, and suppose by induction on n that $A \bigcap X^{(n-1)}$ is open in $X^{(n-1)}$. By definition of the quotient topology on $X^{(n)}$ and since $f_{\sigma}^{-1}(S)$ is open in $B_{\sigma}^n$ for each $\sigma$, $S \bigcap X^{(n)}$ is open in $K^{(n)}$.

My problem is to understand the bold part. I don't know how to use the definition there. I would be glad to have your explanation.

In the lecture we defined a CW-complex in the following way: $K^{(0)}$ is a descrete set of points. These points are called 0-cells. If $K^{(n-1)}$ has been defined, let $\lbrace f_{\partial \sigma} \rbrace_{\sigma \in I_n}$ be a collection of maps $f_{\partial \sigma}:S^{n-1} \to K^{(n-1)}$. For each $\sigma \in I_n$ pick a copy $B_{\sigma}^n$ of $B^n$. $\partial B_{\sigma}^n=S^{n-1}_{\sigma}$. Put $Y=\bigsqcup_{\sigma \in I_n}B_{\sigma}^n$ and $\partial Y=\bigsqcup_{\sigma \in I_n}S_{\sigma}^{n-1}.$Put the maps $f_{\partial \sigma}$ together to get a map $f_{\partial}:\partial Y \to K^{(n-1)}$. Then define $K^{(n)}=K^{(n-1)} \bigcup_{f_{\partial}}Y$

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$K^n$ is a quotient of the disjoint union $K^{n-1}\bigsqcup_{σ\in I_n}B^n_σ$. To check that $S$ is open in $K^n$ it suffices to check that it is open in $K^{n-1}$ (assumed by induction) and that its preimages $f^{-1}_σ[S]$ are each open in $B^n_σ$ for all $σ\in I_n$

Have a closer look at $K^n$. It is defined as $K^{n-1}\cup _{f_\partial}Y$. This notation is used for so-called adjunction spaces. In general, if you have a space $B$, a space $X$ and a subspace $A\subseteq X$ with a map $f:A\to B$, then the adjunction space $B\cup_f X$ is defined as the quotient $B\sqcup X/\sim$ where we identify $a\in A$ with $f(a)∈B$. Let us call the quotient map $q:B⊔X→B∪_f X$, and $\bar i:B→B∪_fX$ and $\bar f:X→B∪_fX$ the restrictions of $q$ to these disjoint components, so $\bar f$ is the composition $X\hookrightarrow X⊔B\twoheadrightarrow B∪_fX$. The map $\bar i$ is actually an embedding, so we can say $B⊆B∪_fX$. By the definition of the quotient topology, in order to show that $S⊆B∪_fX$ is open, it suffices to show that $\bar i^{-1}[S]=S\cap B$ and $\bar f^{-1}[S]$ are open.

In your case this amounts to showing that $S\cap K^{n-1}$ and $\overline{f_∂}^{-1}[S]$ are open. But what is $\overline{f_∂}$? It is the composition $⨆_{σ∈I_n}B_σ^n↪K^{n-1}⨆_{σ∈I_n}B_σ^n↠K^{n−1}⋃_{f_∂}Y$, so on the individual $B_σ^n$ it is the characteristic map $f_σ$.

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  • $\begingroup$ I still didn't understand it. Probably because I haven't seen many quotient spaces yet. Why does it suffice to know, that just the preimages $f^{-1}_{\sigma}(S)$ are open in $B_\sigma^n$? Are the $f_\sigma$ something like the quotient maps? $\endgroup$ – Rungo Oct 27 '13 at 6:15
  • $\begingroup$ @Rungo: See my edited post and feel free to ask if it's still unclear. $\endgroup$ – Stefan Hamcke Oct 27 '13 at 15:26

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