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I want to prove that the suspension $\Sigma X$ of a CW-complex $X$ is a CW-complex, buy I'm starting with CW-complexes and I don't have a clue of how start, so I'd appreciate any help. Thanks.

$$\Sigma X=\frac{X \times I}{X \times \{0,1\}}$$

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    $\begingroup$ Are you required to use this definition of the suspension? I prefer to think of it as connecting every point of the space $X$ to a point "above" and a point "below" by a straight line. $\endgroup$ – N. Owad Jul 10 '14 at 15:31
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    $\begingroup$ No, I'm not required. But I was trying to see something like: $X$ is a CW-complex and $I$ too, so $X\times I$ is also a CW-complex...and so on, but I don't know how to continue of even if this idea would drive me to solution. $\endgroup$ – user148502 Jul 10 '14 at 15:36
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    $\begingroup$ Try it with a single $n$-cell. See what the suspension is. Then try to generalize to a complex. $\endgroup$ – N. Owad Jul 10 '14 at 15:40
  • $\begingroup$ @N.Owad If $e^n=n-cell$, by definition $e^n \cong int(D^n)\cong S^{n-1}$, so $\Sigma e^n \cong \Sigma S^{n-1} \cong S^n$. If $X$ is a CW-complex, $X= \bigcup_{n \in \mathbb{N}} X^{(n)}$, st: $X^{(0)}$ is a discrete set of points; $X^{(n)}=X^{(n-1)}\cup_{f_{\alpha}} \coprod_{\alpha \in \Lambda} e^n_{\alpha}$ and $A\subset X$ is closed in $X$ iff $A \cap X^{(n)}$ is closed in $X^{(n)}, \forall n$. So, studying $\Sigma X$, we see $\Sigma X^{(0)}=S^0=p$; $\Sigma X^{(n)}=X^{(n-1)}\cup_f \coprod S^n$ and $A\subset X$ is closed in $X$ iff $A \cap X^{(n)}$ is closed in $X^{(n)}$, where I'm not right? $\endgroup$ – user148502 Jul 10 '14 at 16:11
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    $\begingroup$ In your first line of the comment above, $int(D^n)\cong S^{n-1}$ seems to be a problem... Think about what you have there. $\endgroup$ – N. Owad Jul 10 '14 at 16:27
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It is a general result (that I invite you to prove) that the product of CW-complexes (one of which is finite) is still a CW-complex, using the product of the cells involved. Hence $X \times I$ is a CW-complex. Moreover, $X \times \{0,1\}$ is a subcomplex thereof, so that $(X \times I,X \times \{0,1\})$ is a relative subcomplex (see this http://www.math.uchicago.edu/~may/CONCISE/ConciseRevised.pdf , at page 73-74): this implies that $\Sigma X=(X\times I) /(X \times \{0,1\})$ is a CW-complex.

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