Let $D$ and $D^\prime$ be bounded domains in the complex plane, and let $f: \overline {D} \rightarrow \Bbb C $ be continuous function that is analytic in $D$. Under the assumptions that $f(D)$ is subset of $D^\prime$ and that $f(\partial D)$ is subset of $\partial D^\prime$, verify that $f(D)$ = $D^\prime$, $f(\partial D)=\partial D^\prime$, and $f(\overline{D})=\overline {D}^\prime$.
My work: Assume that $G=f(D) \ne D^\prime $. Prove by contradiction will work here but i could not get it.
Claim: $D' \subset (D'\setminus f(\overline D)) \cup f(D)$. To see this let $z \in D'$ and suppose $z \notin D'\setminus f(\overline D)$. Then $z \in f(\overline D)=f(D) \cup f(\partial D)$ so $z \in f(D)$ or $z \in f(\partial D)$. But the second possibility is ruled out by the condition $f(\partial D) \subset \partial D'$. [ $D'$ is open so it does not intersect its boundary]. Hence $z\in f(D)$ and this proves the claim. Since $D'$ is connected the claim implies that $D'\setminus f(\overline D)$ is empty. [ Note that $D'\setminus f(\overline D)$ and $f(D)$ are disjoint open sets]. Hence $D' \subset f(\overline D)$. But then $D' \subset f(\overline D)\subset \overline {f(D)}$ by continuity of $f$. This gives $D' \subset f(\overline D)\subset \overline {f(D)}\subset \overline {D'}$. taking closure throughout we get $f(\overline D)=\overline {D'}$. An elementary set theoretic argument now gives $f(D)=D'$ and $f(\partial D)=\partial D'$