A continuous and injective map $f: \mathbb{S}^2 \rightarrow \mathbb{S}^2$ is a homeomorphism

Question

I found this question talking about covering spaces so I am trying to prove that f is a covering map. This give me the surjectivity I need to prove the question since the sphere is compact and $T_2$. Any idea?

Answer 1

First, let's consider the following result: $$\underline{\mathbf{Borsuk-Ulam\: \: theorem \:(n = 2)}}: \: \: f: \mathbb{S}^2\rightarrow \mathbb{R}^2 \: \: \text{continuous} \Rightarrow \exists x \mid f(x) = f(-x).$$ To show that $f$ is bijective we have to prove that $f$ is surjective (because $f$ is already injective). Suppose $f$ isn't surjective, then: $$ \exists p \mid f(\mathbb{S}^2) \subset \mathbb{S}^2 \setminus \left \{ p \right \} \cong \mathbb{R}^2 \overset{B-U}{\Rightarrow} \exists x \mid f(x) = f(-x) \Rightarrow f \: \: \text{not injective !!} $$ So $f$ is bijective. The last thing we need to show is that $f^{-1}$ is continuous. Notice that if $f$ is bijective this is equivalent to show that $f$ is a closed map. $$ C\subset \mathbb{S}^2 \: \: \text{closed} \overset{\mathbb{S}^2 \: \text{compact}}{\Rightarrow} C \: \: \text{compact} \Rightarrow f(C) \: \: \text{compact} \overset{\mathbb{S}^2 \: \text{Hausdorff}}{\Rightarrow} f(C) \: \: \text{closed}. \: \square$$

Answer 2

Assume $f$ is not onto, say it avoids $N\in\Bbb S^2$. Then by stereographic projection with $N$ as north pole, you obtain a continuous injective map $g\colon\Bbb S^2\to\Bbb R^2$. Then consider $h\colon \Bbb S^2\to \Bbb R^2\setminus\{0\}$, $x\mapsto g(x)-g(-x)$. Let $a\in\Bbb S^2$ and $\gamma\colon[0,1]\to\Bbb S^2$ a path with $\gamma(0)=a$, $\gamma(1)=-a$. Define $\tilde\gamma\colon [0,1]\to \Bbb S^2$ as $$\tilde\gamma(t)=\begin{cases}\gamma(2t)&0\le t\le \frac12\\ -\gamma(2t-1)&\frac12\le t\le 1\end{cases} $$ Then you can show that $h\circ\tilde\gamma$ has odd (hence non-zero) winding number around $0$, hence is not contractible in $\Bbb R^2\setminus\{0\}$. But $\tilde\gamma$ is contractible in $\Bbb S^2$, contradiction.

Answer 3

Sphere is compact and $T_2$, a continuous map from a compact space to a $T_2$ one is closed. Closed bijective maps are open.