# $S^n$ homeomorphic to $I^n/\partial(I^n)$

Let $S^n$ the unit sphere in $\mathbb{R}^{n+1}$ and $I=[0,1]$ the unit interval.

By using polar and spherical coordinates, we can show that

$$S^1\text{ is homeomorphic to }I/\partial I,\text{ and}$$ $$S^2\text{ is homeomorphic to }I^2/\partial (I^2)$$ (with the quotient topology).

But what about higher dimensions ? Do we have $S^n\simeq I^n/\partial(I^n)$ ? The usage of trigonometric function can't be generalized, can it ?

Same question for $S^{n}\simeq \partial (I^{n+1})$: we can construct homeomorphisms in $\mathbb{R}^2$ and $\mathbb{R}^3$ through projections on the faces, but the argument seems to be very hard (or a lot of work) to generalize.

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The structure of trigonometric functions can be generalized (see wiki), but it's not the nicest way to do this problem. – Gerben Oct 2 '11 at 10:06

## 2 Answers

$S^n$ is (homeomorphic to) the one-point compactification of $\mathbb{R}^n$, $\mathbb{R}^n$ is homeomorphic to $(0,1)^n$, the interior of $I^n$, and $I^n/\partial(I^n)$ is homeomorphic to the one-point compactification of $(0,1)^n$. Put the pieces together, and you get $S^n \cong I^n/\partial(I^n)$.

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You can introduce the structure of CW-complex on $I^n$ so that the interior is contained in one $n$-dimensional cell, and then use the fact that $I^n / \partial I^n$ has just two cells: one $n$-dimensional, and one $0$-dimensional, and there can be only one such CW-complex. See Hatcher's Algebraic Topology for more.

About the second question: use the fact that $I^n \cong D^n$, where $D^n$ is the closed $n$-dimensional disk (hint: they are both closed balls in some norms on $\mathbb{R}^n$), and that $\partial D^n = S^{n-1}$.

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