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I was trying to prove some homeomorphism. But I´m not sure of all. This is the problem There are homeomorphic this two sets, with the subspace topology $$ \eqalign{ & S_n ^ + = \left\{ {\left( {x,y,z} \right) \in R^3 :x^n + y^n + z^n = 1\,\,\,\,\,\,\,\,z > 0} \right\} \cr & D_2 = \left\{ {\left( {x,y} \right) \in R^2 :x^2 + y^2 < 1\,\,\,} \right\} \cr} $$ Not sure, so please be grateful if I am wrong T_T. But for example with $n$ even. I can clear I z, which is determined by positive only with $x$ and $y$. Then it seems more natural homeomorphism defined as the projection, as is clearly bijection, we also know that it is open and continuous mapping. Then it is homeomorphism. In the case of $n$ odd I´m not sure, the problem here, is that I can take any value of R, with $x$ or $y$, and the projection , wont be useful.

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You should read some good tutorial on LaTeX to learn of the sensible ways of adding space in equations! :) – Mariano Suárez-Alvarez Aug 30 '11 at 3:33
up vote 2 down vote accepted

The two sets are not homeomorphic; they have different cutsets, and cutsets are preserved by homeomorphisms. Specifically, removing a point from $D_2$ disconnects it, but removing a point from $S^n_+$ does not disconnect it. The idea here is that if $h:X\rightarrow Y$ is a homeomorphism, so is its restriction $h':(X-{pt})\rightarrow (Y-h(pt))$, but , in this case, $X-{pt}$ is connected, but $Y-h(pt)$ is not, which cannot happen between homeomorphic spaces.

EDIT: As Jacob pointed out, I (again ) solved the wrong problem of showing $D^2_+$ and $S^1$ were not homeomorphic.

Here is a solution for the actual question:

EDIT#2: Both spaces are seen to be homeomorphic to $\mathbb R^2$: for $S^n_+$, we have that it is a compact (closed, as the inverse image of {0} of the continuous function $f(x,y,z)=x^n+y^n+z^n-1 $, and clearly bounded, and Daniel's projection map is a continuous bijection, so we have a continuous bijection between compact and Hausdorff, which is a homeomorphism. We also have that $D_2$ is homeomorphic to $\mathbb R^2$, by , e.g., a variant of the $tan^{-1}$ argument used to show that any open interval (a,b) is homeo. to $\mathbb R^1$. The two spaces are then homeomorphic to each other.

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$D_2$ is the open unit disk and is homeomorphic to $\mathbb R^2$ which certainly doesn't have any cutpoints. – JSchlather Aug 30 '11 at 3:59
Right; I misread something, let me check. – gary Aug 30 '11 at 4:04
I think we can use the fact that the two are manifolds, and $S^n+$ has (manifold) boundary, but $D^2$ does not have manifold boundary. – gary Aug 30 '11 at 4:09
What's the boundary of $S^n_+$? – JSchlather Aug 30 '11 at 4:14
OK, I misread that again. Ultimately, $D^2$ is contractible, but $S^n_+$ is not; $D^2-{pt}$ retracts to S^1, and $S^n_+-{pt}$ is homeo. to $\mathbb R^n$ . – gary Aug 30 '11 at 4:29

So I think you're just about there. Prove the following two things and I think you'll see how to finish the problem.

If $n$ is a positive integer then $\pi: S^n_+ \rightarrow \mathbb R^2$ is an embedding where $\pi(x,y,z)=(x,y)$.

Show $\pi(S^n_+) \approx \mathbb R^2$.

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