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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!

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  • $\begingroup$ $\mathbb{S}^n$ 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
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    $\begingroup$ 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
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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).

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    $\begingroup$ 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

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