# In the process of proving 'Invariance of Domain'

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.

• 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 $B$. – Ted Shifrin Dec 16 '19 at 2:13
• Thanks for your help – HooMun Dec 16 '19 at 2:18