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!



a space isomorphic to $S^{p+q}$
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).


