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I want to show that the following map induces a homeomorphism between the two structures.

$\require{AMScd}$ \begin{CD} \mathbb{B}^{n} @>f(x_1,\cdots , x_n) = (x_1 , \cdots x_n , \sqrt{1- \sum_{i=1}^n(x_i)^2})>> \mathbb{S}^n @>\pi_1(x) = cl(\{x\}) >> \mathbb{S}^{n} / \ \sim_2 \\ @VV\pi_2(x) = cl(x)V \\ \mathbb{B}^{n} / \sim_1 \\ \end{CD}

where $\sim_1$ signifies identifying the antipodal points of the boundary circle of $\mathbb{B}^n$ and $\mathbb{S}^n / \sim_2$ signifies identifying the antipodal points of $\mathbb{S}^n$.

I want to show that $\mathbb{S}^n / \sim_2$ and $\mathbb{B}^n / \sim_1$ are homeomorphic using the theorem,

$\require{AMScd}$ \begin{CD} \mathbb{X} @>g(x)>> Y \\ @VV p V \\ \ \mathbb{X} / \sim \\ \end{CD}

where $g(x)$ is an identification map and $p$ is a projection map.

The part where I am stuck at is:

1)Showing that $\pi_1 \circ f (x) $ is a surjective map.

2)Is showing that $\pi_1 \circ f(x)$ surjective enough? Or is there anything else we have to show to prove that the two maps are homeomorphic?

Edit $1$:Let $(x_1, \cdots , x_{n+1}) \in \mathbb{S}^n/\sim_2$ then,

case $1:$ Assume that $x_{n+1} > 0$ then $\pi_1^{-1}((x_1,\cdots ,x_{n+1})) = (x_1, \cdots ,x_{n+1})$ where $\sum_{i=1}^{n+1}(x_i)^2 = 1$.

$f^{-1}(x_1,\cdots,x_{n+1}) = (x_1,\cdots x_n)$ s.t. $\sum_{i=1}^n(x_i)^2 < 1$.

Then $$\pi_2(x_1, \cdots x_n) = (x_1 , \cdots x_n)$$

case $2:$ Assume that $x_{n+1} = 0$ then $(x_1,\cdots x_{n+1})$ is a point on the equator of the sphere. Then, $\pi_1^ {-1} (x_1,\cdots x_{n},0)= \{[x],[x]'\} $ where $[x]= (x_1, \cdots ,x_{n+1})$ and $[x]'$ is the antipodal point of $[x]$.

Also $\sum_{i=1}^n(x_i)^2 = 1$ . I am stuck in this part.

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  • $\begingroup$ Oh sorry I have fixed it. $\endgroup$
    – Antimony
    Jul 29, 2021 at 13:32
  • $\begingroup$ The maps make more sense now. $\endgroup$ Jul 29, 2021 at 13:40

1 Answer 1

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$\pi_ 1\circ f$ is surjective because every class in the boundary (for $\sim_2$) has a representative with last coordinate $\ge 0$ which will be in the image of $f$ in an obvious way. You also have to show that the only way we can have $\sim_1$ images equal for distinct points is when the $f$ images are $\sim_2$ equivalent too, to have well-definedness and injectivity of the completing map. Compactness will then do the rest.

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  • $\begingroup$ Can you go through my edit.I am having a problem to continue from there $\endgroup$
    – Antimony
    Jul 29, 2021 at 14:52

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