# Showing that the corresponding two spaces are homeomorphic.

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.

• Oh sorry I have fixed it. Jul 29, 2021 at 13:32
• The maps make more sense now. Jul 29, 2021 at 13:40

$$\pi_ 1\circ f$$ 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.