I have been thinking about various ways to construct an $n$-sphere. Starting with $S^2$, we can construct it by taking two disks, lifting the "meat" of the disks into a third dimension and then identifying the boundaries. My intuition tells me this should be true for $S^3$ as well by taking two 3-balls and identifying the two boundaries. Should this be true for $S^n$? I.e. can we construct $S^n$ by taking two $n$-balls and identifying the boundaries? Some commentary on these constructions would be welcome as well!


1 Answer 1


Yes, that's the way to do it. The two $n$-balls are the upper and lower hemispheres of $S^n$. If you think of $S^n$ as the set of point in $\mathbb R^{n+1}$ with unit norm, then you can decompose this into two sets according to whether the first coordinate is $\geq 0$ or $\leq 0$. Each of these sets is homeomorphic to an $n$-ball by forgetting the first coordinate, giving a projection into $\mathbb R^n$ whose image is all points of norm $\leq 1$.

  • $\begingroup$ @TylerHG: I don't quite understand your question about the circle. Are you asking for topological reasons why the circle is irksome? Or are you just asking how this construction works for a circle? (The same way.) $\endgroup$ Apr 22, 2016 at 18:43
  • $\begingroup$ Sorry ;). I see it now...forgot that a 1-ball is a segment! Kept thinking of a 0-sphere for some stupid reason $\endgroup$ Apr 22, 2016 at 21:13

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