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I have seen a similar questions here but i want to know is the idea of the next proof is right

We know that $S^n$ has a CW structure given by $1$ $0$-cells $e^0_{\alpha}$ and $1$ $n$-cell $e^n_{\alpha}$ and the same for $S^m$ has a CW structure given by $1$ $0$-cells $e^0_{\beta}$ and $1$ $m$-cell $e^m_{\beta}$

and then $S^n \times S^m$ has the next cells $e^0_{\alpha} \times e^0_{\beta}$, $e^n_{\alpha} \times e^0_{\beta}$ $e^n_{\alpha} \times e^m_{\beta}$ and $e^0_{\alpha} \times e^m_{\beta}$

but the smash product is defined to be $S^n \times S^m /S^n \vee S^m$ and then $e^0_{\alpha} \times e^0_{\beta}$, $e^n_{\alpha} \times e^0_{\beta}$ and $e^0_{\alpha} \times e^m_{\beta}$ are "colapsed" to one point so we have a point and a $e^n_{\alpha} \times e^m_{\beta}$ cell and then is homeomorphic to $S^{n+m}$

Duong Ngo
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Diego Lopes
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  • The boundary of $e_{\alpha} ^n\times e_{\beta} ^n$ is collapsed to a point, and this gives a sphere . – Thomas Mar 01 '25 at 13:13
  • @JohnPalmieri thanks but how i wrote i have seen that answers but im question about if the idea of this proof is right because it doesnt say any similar in the link you post – Diego Lopes Mar 01 '25 at 18:08
  • @Thomas thanks Then I think that is a correct way to prove it – Diego Lopes Mar 01 '25 at 18:13

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