# Show that $X^{*}$ with the quotient topology is homeomorphic to $\mathcal{S}^{2}$.

Consider $\mathbb{R}$ with the usual topology, $\mathbb{R}^{2}$ and $\mathbb{R}^{3}$ with the product topology. Also, consider $X=[0,1]\times [0,1]$ as a subspace of $\mathbb{R}^{2}$ and $\mathcal{S}^{2}=\{(x,y,z)\mid x^{2}+y^{2}+z^{2}=1\}$ as a subspace of $\mathbb{R}^{3}$. On $X$ consider the following partition, $$X^{*}=(0,1)\times(0,1)\cup\{(0,y),(1,y)\mid 0<y<1\}\cup\{(x,0)\mid0\leq x\leq 1\}\cup\{(x,1)\mid 0\leq x\leq1\}$$ Show that $X^{*}$ with the quotient topology is homeomorphic to $\mathcal{S}^{2}$.

Can give me any hint for this problem, I though define a function define in terms of sine and cosine, is the first time I see quotient topology. Thanks!

• @positrón0802 $X$ is partitioned into the union of the two vertical segments, the two horizontal segments, and the singletons of the open square.
– Pedro
Jul 30, 2017 at 7:41

Consider the map $F : [0,1]\times [-1,1]\to S^1\times [-1,1]$ such that $F(t,x)=(e^{2\pi it},x)$. Note that $F(1,x) = F(0,x)$ for every $x\in [0,1]$. Now consider the map $G : S^1\times [-1,1] \to S^2$ such that $G(z,x) = (\sqrt{1-x^2}\,z,x)\in S^2$. In this way you obtain a map $H = GF : [0,1] \times [-1,1]\to S^2$ with the property that
$$H(1,x) = H(0,x) \text{ for each x\in [-1,1]}$$
$$H(t,1) = (0,0,1) \text{ for each t\in [0,1]}$$
$$H(t,-1) = (0,0,-1) \text{ for each t\in [0,1]}$$
This means $H$ factors through the quotient of $[0,1]\times [-1,1]$ where we identify $0\times [-1,1]$ with $1\times [-1,1]$ and where we collapse $[0,1]\times 1$ and $[0,1]\times -1$ to a point.
If you prove this map, which is continuous, is a bijection, and that the quotient space obtained above is compact you are done, because $S^2$ is Hausdorff, and any continuous bijection from a compact space to a Hausdorff space is a homeomorphism.