Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

1 Answer

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 $S^n$.

share|improve this answer
    
thank you again –  Juan S Mar 22 '11 at 6:29
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.