# The restriction of f to U is injective or surjective or both…?

Let $$D$$ be the open unit disc in the complex plane and $$U=D\setminus \{-1/2 ,1/2\}$$.

Also let $$H_{1}=\{f:D\rightarrow\mathbb{C}\mid \text{f is holomorphic and bounded} \}$$ $$H_{2}=\{f:U\rightarrow\mathbb{C}\mid \text{f is holomorphic and bounded} \}$$

Then the map $$r:H_{1} \rightarrow H_{2}$$ given by $$r(f)=f|_{U}$$, the restriction of $$f$$ to $$U$$, is injective or bijective or both...

My Efforts: I tried to solve this , using the concept of restriction map but didn't get any hint.Please give some idea...

• Use \{ and \} for braces, and \text{ some text } for text inside math. – metamorphy Nov 1 '19 at 9:00

## 1 Answer

If $$r(f)=r(g)$$ then $$f(z)=g(z)$$ except when $$z \in \{-\frac 1 2 , \frac 1 2\}$$. But then continuity implies $$f(z)=g(z)$$ for all $$z$$ so $$r$$ is injective. It is surjective because any bounded holomorphic function on $$U$$ can be extended to a holomorphic function on $$D$$ since it has removable singularities at $$\pm \frac 1 2$$.