Let $f\in\mathcal{H}(\mathbb{D})$ (i.e. it is a holomorphic function on the open unit disc) and $\gamma$ a (open) circular arc $\in \partial\mathbb{D}$ such that $f\in\mathcal{C}^0(\mathbb{D}\cup\gamma$); $f(\gamma)$ is an analytic curve. Prove that $f$ has an analytic extension across $\gamma$
My attempt: Since $f(\gamma)$ is an analytic curve, we know that there exists a conformal map $\varphi:A\to \mathbb{C}; \varphi((0,1))=f(\gamma)$. Then, considering $g(z):=\varphi^{-1}\circ f(z)$ (on $f^{-1}(\varphi(A))$), it satisfies the hypotesis of Schwarz's reflection principle for the circle, and so it is extendable.
Applying $\varphi$ to the extension, we obtain an extension of $f$
Is my attempt right? Are there other way of proving it?
I am aware of the reflection principle for analytic arcs, but I am not sure on how to prove that $f(\mathbb{D})$ lies on a side of $f(\gamma)$ and so I cannot use it.