A question on mapping inside unit disc For an analytic function $f:\bar D→\bar D$ where $\bar D$ is the closed unit disc centered at origin.Suppose $\bar D=D\cup$$\delta D$, where $\delta D$ is the boundary of open disc $D$ and $f$ is onto, then is it true that $f(\delta D)=\delta f(D)$ i.e. boundary of domain maps onto the boundary of image?


*

*My attempt: I think the answer is yes. I took $f(z)$=$(z-a)/(1-$$\bar a z$) where $a\in D$ which gives $|f(exp(i\theta)|=1$.


It follows from the open mapping theorem that no interior point can map to the boundary, so $\partial D$ must be contained in $f(\partial D)$.  For the reverse inclusion, you can note that $g(t)=|f(e^{it})|^2$ is a real-analytic function $\mathbb{R}\to\mathbb{R}$ which must take the value $1$ for uncountably many values of $t$ (since there are uncountably many points in $\partial D$ and each must be in the image $f(\partial D)$).  Thus $g(t)$ must be equal to $1$ everywhere, which says $f(\partial D)\subseteq\partial D$.
