Does a continuous mapping have to map the boundary to the boundary

This question comes from when I read my Complex Analysis book.

On the chapter of Mobius transformation, there is an example:"construct a Mobius transformation that maps $\{\mathrm{Im} (z)>0\}$ onto $\{|z|<1\}$". In the solution, it just starts with "it must maps the real line to the circle".

I think about it and try to prove this . I know that the open mapping theorem can do this, since the theorem suggests that a holomorphic function must map the interior point to inner point, thus it also maps the boundary point to boundary point. But in the book, the open mapping theorem is in the later chapters. And I come up with another proof: since Mobius transformation maps a circline to a circline, it should map the real line to a circle. If it is not the boundary, there would be a contradiction that the image have points both within and outside that circle.

But I come up with another more general question. For a continuous function from $R^2$ to $R^2$, if it maps a region onto a region(say it maps $\{y>0\}$ onto $\{x^2 + y^2 < 1\}$), can we claim it must maps the $x$-axis to the unit circle? Or can we say it maps an interior point to an interior point?

In the one-variable function it is not right, $\sin x$ is a counterexample. But I am not sure if there is any counterexample for the $R^2$ function.

Thank you!

• Möbius transformations are homeomorphisms of the Riemann sphere. Homeomorphisms map boundaries to boundaries. – Daniel Fischer Jan 6 '18 at 18:31

$$f(x+iy) = \frac{(y-1)^2}{1+(y-1)^2}e^{ix}.$$
Then $f$ is continuous on $\mathbb R^2$ and $f$ maps $\{y>0\}$ onto $\{x^2+y^2<1\}.$ But $f(\mathbb R)= \{|w|=1/2\}.$
If $f\colon\mathbb{R}^2\longrightarrow\mathbb{R}^2$ is the null function, then it maps the set $\{z\in\mathbb{C}\,|\,\operatorname{Im}z>0\}$ into the open unit disk, but no real point is mapped to the unit circle.
By the way, suppose that $f(D) = D'$ are open regions where D has compact closure. Then you statement holds, i.e. $f(\partial D) = \partial D'$. Infact if $y \in \partial D'$, take a succession $y_n$ which converges to $y$. Take $x_n$ such that $f(x_n) = y_n$ and extract a convergent subsuccession, say to $x$. Then $f(x) =y$, and $x \in \partial D$, otherwise $x \in D \ \Rightarrow \ y \in D'$ which is disjoint with its boundary.
In your case, this proof still works because mobius functions extends to $S^2$ (adding "infinity" to $R^2$), which is compact (thus every set has compact closure).