We have $S^k \times S^l$ with cellular structure $e^0_1\times e^0_2,~ e^k\times e^0_2,~ e^0_1\times e^l,~ e^k\times e^l.$ Why do first three cells form $S^k \vee S^l$?

I mean, if we assume $k<=l$, at first we have a point, then we must glue boundary of a $k$-cell to that point obtaining sphere $S^k$, but then we need to glue boundary of an $l$-cell to $S^k$ and I do not see why it must all be glued to the same point.

Thank you.

  • 4
    $\begingroup$ Well, the reason why it’s all glued to the point is that that is the definition of the cell structure on the product! The attaching map of a product of cells is the product of the attaching maps of each factor. $\endgroup$ Commented Apr 20, 2020 at 1:02
  • $\begingroup$ @KevinCarlson How is product of continuous maps defined? $\endgroup$
    – Haldot
    Commented Apr 20, 2020 at 9:04

1 Answer 1


Kevin in the comments means that the characteristic map of the product cell is the product of the characteristic maps of the two factors (rather than the attaching map).

Assume that we have attached an $m$-cell to $X$ and and $n$-cell to $Y$ to give us characteristic maps $\Phi:D^m\rightarrow X\cup e^m$ and $\Psi:D^n\rightarrow Y\cup e^n$. The attaching maps are their restrictions to the boundaries $\partial D^m=S^{m-1}$ and $\partial D^n=S^{n-1}$. i.e. $\phi=\Phi|:S^{m-1}\rightarrow X$ and $\psi=\Psi|:S^{n-1}\rightarrow Y$. If you want to think in terms of a pushout square defining the adjunction space, then the attaching and characterstic maps live on opposite sides of the square.

Now the product of the characteristic maps $\Phi,\Psi$ is simply the map $$\Phi\times \Psi:D^m\times D^n\rightarrow (X\cup e^n)\times (Y\cup e^n),\qquad (z_1,z_2)\mapsto (f(z_1),g(z_2)).$$

The point is that there is a homeomorphism $D^m\times D^n\cong D^{m+n}$, which we can see by observing that $D^n\cong I^n$. On the boundary this homeomorphism restricts to a homeomorphism $$\partial(D^m\times D^n)=(\partial D^m\times D^n)\cup(D^m\times \partial D^n)\cong \partial D^{m+n}$$ which we might prefer to write $$(S^{m-1}\times D^n)\cup(D^m\times S^{n-1})\cong S^{m+n-1}.$$

To get the attaching map of the product cell we simply restrict the characteristic map to the boundary. So if $\Phi\times \Psi$ above is the characteristic map, then the attaching map it defines is $(\Phi\times \Psi)|_{\partial D^{m+n}}$. Using the homeomorphism above this map is exactly $$(\phi\times\Psi)\cup(\Phi\times\psi):(S^{m-1}\times D^n)\cup(D^m\times S^{n-1})\rightarrow (X\times (Y\cup e^n))\cup ((X\cup e^m)\times Y).$$ where of course the domain is homeomorphic to $S^{n+m-1}$

The point is that we should get a homeomorphism between the product space $(X\cup e^n)\times (Y\cup e^n)$ and $((X\times (Y\cup e^n))\cup ((X\cup e^m)\times Y))\cup e^{n+m}.$

If we take $X=Y=\ast$, then $X\cup e^m\cong S^m$ and $Y\cup e^n\cong S^n$, and the above argument says that there is a homeomorphism

$$S^m\times S^n\cong ((\ast\times S^n)\cup (S^m\times \ast))\cup e^{n+m}=(S^m\vee S^n)\cup e^{m+n}.$$


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