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Every characteristic function $\Phi_{\beta}^b: E^n_{\beta} \to e^{-n}_{\beta}$ is an identification function.

My book says the following:

This follows from the fact the the CW complex $X$ has $X^n$ a quotient space of $X^{n-1} \sqcup \coprod_{\beta}E^n_{\beta}$.

I am not sure I see how this follows. We know that $\Phi_{\beta} |S^{n-1}_{\beta} = \phi_{\beta}:S^{n-1}_{\beta} \to X^{n-1}$, but how do I use what the book gave me to prove it?

Note: $e^{n}_{\beta}$ is an open $n$-cell, $E$ is the unit ball, and $S$ is the unit sphere. The characteristic function is what I have defined above.

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  • $\begingroup$ Can you define all the notation you're using, and also the terms "characteristic function" and "identification function"? $\endgroup$ – Eric Wofsey Apr 11 '16 at 2:11
  • $\begingroup$ @EricWofsey An identification, $q$, map satisfies two conditions: the function $q: X \to Y$ is surjective and a subset $U$ of $Y$ is open if and only if $q^{-1}(U)$ is open in $X$. $\endgroup$ – user19405892 Apr 11 '16 at 2:15
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I don't see an easy way to prove this using what the books says. But if you know that CW complexes are Hausdorff, you can instead prove it as follows. Let $Y$ be the quotient space of $E^n_\beta$ by the equivalence relation induced by $\Phi^b_\beta$, i.e. $x\sim y$ iff $\Phi^b_\beta(x)=\Phi^b_\beta(y)$. Then $Y$ is compact since $E^n_\beta$ is, and $\Phi^b_\beta$ gives a continuous bijection $f:Y\to \overline{e^n_\beta}$. Since $X$ is Hausdorff, so is $\overline{e^n_\beta}$, and so $f$ is automatically a homeomorphism (it is a continuous bijection from a compact space to a Hausdorff space). But $\Phi^b_\beta$ is just $f$ composed with the quotient map $E^n_\beta\to Y$, so this means $\Phi^b_\beta$ is also a quotient map.

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  • $\begingroup$ I think my book is deriving it from the fact that the $n$-skeleton $X^n = X^{n-1} \cup_{\phi} \coprod_{\beta \in B} E^n_{\beta}$. $\endgroup$ – user19405892 Apr 11 '16 at 2:32
  • $\begingroup$ The problem is that in general the restriction of a quotient map is not a quotient map. In fact, to get the conclusion you want, you have to use something like knowing that $X$ is Hausdorff, because if you let $X^{n-1}$ be an arbitrary space and attach a cell to it by a map from $S^{n-1}$, the resulting map from $E^n$ need not be a quotient map in general. $\endgroup$ – Eric Wofsey Apr 11 '16 at 2:42
  • $\begingroup$ What exactly is a characteristic function here? It isn't the indicator function here right? $\endgroup$ – user19405892 Apr 11 '16 at 21:57
  • $\begingroup$ I assume from context that the "characteristic function" of a cell is the restriction of the quotient map $X^{n-1}\sqcup \coprod E^n_\beta\to X^n$ to $E^n_\beta$. $\endgroup$ – Eric Wofsey Apr 11 '16 at 23:21

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