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Let $Y$ be a Hausdorff space, and let $f:S^{n-1} \to Y$ be continuous. Then $D^n \coprod_f Y$ is called the space obtained from $Y$ by attaching an n-cell (denoted $e^n$) via $f$ and is denoted $Y_f$

The charactersitc map $\Phi$ is the composite:

$D^n \hookrightarrow D^n \coprod Y \to D^n \coprod_f Y \to Y_f$ so that $\Phi:(D^n,S^{n-1}) \to (Y_f,Y)$ is a function of pairs, and $\Phi|S^{n-1}$ is the attaching map.

(See here for $D^n \coprod_f Y$ if this is non-standard).

As a (presumably) gentle introduction to CW-complexes I have the following question:

If $Y$ is a singleton, show that the space obtained from $Y$ by attaching an n-cell is $S^n$, hence $S^n = e^0 \cup e^n$ (where $e^n \simeq D^n - S^{n-1}$).

I am unsure how to approach this problem. If I let $y=(-1,0,\ldots,0)$, then I believe there is a map $\Phi:(D^n,S^{n-1}) \to (S^n, y)$, such that $\Phi|e^n$ is an embedding.

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up vote 1 down vote accepted

One way to proceed is to show that if $X$ is a space, $A\subseteq X$ a subspace, and $f:A\to\{*\}$ is the constant map to a one-point space $\{*\}$, then the result of attaching $X$ to $\{*\}$, $X\sqcup_f\{*\}$ is the same thing as the result of collapsing $A$ to a point, $X/A$.

Next, show that the collapsed space $D^n/S^{n-1}$ is an $S^n$.

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thank you again – Juan S Mar 22 '11 at 6:29

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