If $U$ is an open subset of euclidean plane $R^2$, and $f: U \rightarrow R^2$ is continuous and injective, then $f(U)$ is open in $S^2$ and the inverse function is continuous.

This is Theorem 62.3 in Topology, Munkres. According to this book,

Step 1. We show that if $B$ is any closed ball in $R^2$ contained in $U$, then $f(B)$ does not separate $S^2$.

Let $a$ and $b$ be two points of $S^2-f(B)$. Because the identity map $i: B \rightarrow B$ is nulhomotopic, the map $h:B \rightarrow S^2 - a - b$ obtained by restricting $f$ is nulhomotopic.

But, I coudln't understand how restriction of $f$ be nulhomotipic.

  • $\begingroup$ Isn't this the same argument as appears at the beginning of the proof of Lemma 62.2? Because $B$ is compact (and Hausdorff) and $f$ is injective, $f(B)$ is homeomorphic to $B$. $\endgroup$ – Ted Shifrin Dec 16 '19 at 2:13
  • $\begingroup$ Thanks for your help $\endgroup$ – HooMun Dec 16 '19 at 2:18

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