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Let $f$ be analytic function defined on unit disc such that $f$ is univalent on each circle $C_r = \{z : |z|=r \}$ for each $r<1$. Does it follow that $f$ is univalent in the unit disc? Say $f$ is not univalent, then there will be two points $z_1$ and $z_2$ such that $f(z_1) = f(z_2)$ and $|z_1| \ne |z_2|.$ We know that if $0 < r < 1,$ then $f(rz)$ is univalent. So here we may assume that $|z_1| < |z_2|.$ I tried to consider the point $f(z_1)$ and $f(z_2 |z_1|/|z_2|)$ and they cannot be equal, as they lie on the same circle. But I am not able to get any contradiction from here. If I take some examples and check the assertion seems to be true but unable to prove it. Are there counter examples to this?

Eloon_Mask_P
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    The answer is yes by the argument principle (the equation ($f(z)=w$ has precisely $\chi (J_r,w)$ solutions where $J_r=f(D_r)$ analytic Jordan curve) so it has $0$ if $w$ outside and $1$ inside ) – Conrad Apr 30 '23 at 16:41
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    Even more is true - if $f$ is analytic in the open disc and continuous only on the boundary, then if $f$ is injective on the boundary, it is univalent in the open disc too and vice versa (this last part is Caratheodory theorem and first follows from Jordan curve theorem) – Conrad Apr 30 '23 at 16:52

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