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Let $S^n$ denote the $n$-sphere and $D^n$ denote the $n$-disk (of course, $\partial D^{n+1}\cong S^n$). Then $S^n\times D^{m+1}$ and $D^{n+1}\times S^m$ both have boundary $S^n\times S^m$. The exercise I am trying to solve asks to prove that $S^{n+m+1}$ can be decomposed as the union of these two spaces along their boundaries.

It seems to me that it should be a classic (perhaps easy) result but I am quite lost on where to start. I have been trying to come up with an explicit homeomorphism but I have not found any.

There is a related question for a special case of this result, but it suggests using a Morse function, which I rather avoid if possible.

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Mauricio Tec
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3 Answers3

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Use the fact that the n-disks disks are product spaces:

$D^n \times D^m \simeq D^{n+m}$. Then $\partial D^{n+m+2}=S^{n+m+1}=\partial(D^{n+1}\times D^{m+1})=S^n \times D^{m+1} \cup D^{n+1}\times S^m$

and you're done.

user203864
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Let $S^{n+m+1}$ be given by

$$S^{n+m+1} = \{(\vec x, \vec y): |\vec x|^2 + |\vec y|^2 = 1\},$$

where $\vec x = (x_1, \cdots, x_{n+1})$ and $\vec y = (y_1, \cdots, y_{m+1})$. Then $S^{n+m+1}$ decomposes into two parts:

$$A = \{ |\vec x| \ge |\vec y| \} \cup B = \{ |\vec x| \le |\vec y|\}$$

which shares a common boundary $\{ |\vec x| = |\vec y| = \frac 12\}$. Note that the common boundary is homeomorphic to $\mathbb S^n \times \mathbb S^m$.

Now define a continuous map: $F : \mathbb S^n \times \mathbb D ^{m+1} \to A$ by

$$F(\vec u, \vec v) = \frac{1}{\sqrt{1 + |\vec v|^2}} \big( \vec u, \vec v\big) \in \mathbb S^{n+m+1}$$

first of all, as $1= |\vec u|\ge |\vec v|$, we have $F(\vec u, \vec v) \in A$. Now we show that it is bijective:

If $F(\vec u_1, \vec v_1 ) = F(\vec u_2, \vec v_2)$, then $\vec u_1 = C \vec u_2$ for some scalar $C$, which implies that $\vec u_1 = \vec u_2$ as they both lie in $\mathbb S^{n+1}$. This also shows that $|\vec v_1| = |\vec v_2|$ and so $\vec v_1 = \vec v_2$. Thus $F$ is injective.

To show that it is surjective, note that for all $(\vec x, \vec y) \in A$, we have

$$F\bigg(\sqrt{1 + \frac{|\vec y|^2 }{|\vec x|^2}} \vec x , \frac{\vec y}{|\vec x|}\bigg) = (\vec x, \vec y)$$

Thus $F$ is surjective. As $\mathbb S^{n+1} \times \mathbb D^{m+1}$ is compact and $A$ is Hausdorff, $F$ is a homeomorphism.

One can similarly construct a homeomorphism $\mathbb D^{n+1} \times \mathbb S^{m+1} \to B$.

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$S^{n+m+1}$ is the one point compactification of $\mathbb{R}^{n+m+1}$. Now decompose $\mathbb{R}^{n+m+1}$ into $S^n \times D^{m+1}$ and its complement. Note that the complement has boundary the hollow torus $S^n \times S^m$. In the compactification $S^{n+m+1}$, this complement becomes the the second solid torus $D^n\times S^m$.

This is hard to visualize. A nice explanation for $S^3$ is given in page $48$ of Hatcher's book.

ChesterX
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