I've asked this question few days ago in MathStackExchange, but there was no result. So, I decided to ask it there.

In one of the paper I have met that $$\mathbb{S}^{p+q} \cong \mathbb{S}^p \times \mathbb{R}^q \cup \mathbb{S}^{q-1}$$

I got stuck here, don't know how to prove that.

Are there any references for this isomorphism?

Thanks in advance!

  • $\begingroup$ $\mathbb{S}^n$ here is the sphere, I suppose? What does $\cong$ mean here? My first guess would be "homeomorphic" but that clearly isn't right... Can you cite the paper where this appears? $\endgroup$ – Nate Eldredge Jul 23 '16 at 2:57
  • $\begingroup$ @NateEldredge yes, that is the sphere. I found that in Hambleton-Pederson paper, Compactifying infinite group actions $\endgroup$ – Jane Jul 23 '16 at 3:28
  • 3
    $\begingroup$ It looks like there is a misprint in the formula and the exponent $p-1$ should be $q-1$. For example, we have $S^3 - S^0 = S^2 \times {\mathbb R}$ and $S^3 - S^1 = S^1 \times {\mathbb R}^2$. $\endgroup$ – Allen Hatcher Jul 23 '16 at 4:17
  • $\begingroup$ @AllenHatcher you are right, Dr. Hatcher. I updated the formula above. P.S. your Algebraic Topology book is awesome :) $\endgroup$ – Jane Jul 23 '16 at 4:45

This formula is a version of another one, which I find more elegant: $$ S^a * S^b \simeq S^{a+b+1} $$ where $a,b$ are non-negative integers, and $*$ denotes the topological join (https://en.wikipedia.org/wiki/Join_(topology)): $X*Y$ is obtained by taking a disjoint union of $X$ and $Y$, and for each pair $(x,y)\in X\times Y$ to glue a segment from $x$ to $y$. The topology is obtained by deciding that segments with both endpoints close are close.

The above formula is easy to prove when $b=0$, since one then gets a suspension of $S^a$. Since the join is associative (at least for locally Hausdorff spaces I think), the above formula is easily proved by induction on $b$.

Now, to get your formula, consider $S^{p+q}\simeq S^p * S^{q-1}$ and remove the copy of $S^{q-1}$ from the construction of the join. You are left with the points of $S^q$, each of which comes with a bunch of open intervals, one for each point of the removed $S^{q-1}$. This bunch makes a $S^{q-1}\times (0,+\infty)$, which together with the considered $S^p$ point makes a $\mathbb{R}^q$. So $S^{p+q}\setminus S^{q-1}$ is $S^p\times \mathbb{R}^q$ (if you remove the right $S^{q-1}$, of course).

  • 2
    $\begingroup$ A direct way to prove that formula is using the map $S^p\star S^q\to S^{p+q+1}$ given $(x,t,y)\in \mathbb{R}^{p+1}\times [0,1]\times \mathbb{R}^{q+1}\mapsto (x\cos t\pi/2, y \sin t\pi/2) \in \mathbb{R}^{p+q+2}$ $\endgroup$ – Denis Nardin Jul 23 '16 at 12:45
  • $\begingroup$ @DenisNardin Thank you for the valuable comment! $\endgroup$ – Jane Jul 23 '16 at 16:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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