Given $A\subset\mathbb{R}^n$, not necessarily open, and $f:A\rightarrow\mathbb{R}^m$ a function with $\text{graph}(f)\subset\mathbb{R}^{n+m}$ an analytic submanifold, by which I mean for all $x\in\text{graph}(f)$ there exists open $B\ni x$, $\Omega\ni 0$ in $\mathbb{R}^{n+m}$ an an analytic diffeomorphism $\varphi:B\rightarrow\Omega$ such that $\varphi(\text{graph}(f)\cap B)=(\mathbb{R}^{\text{dim}(\text{graph}(f)}\times\{0\})\cap \Omega$, can I conclude that $f$ is an analytic function of $x$? I don't think $A$ being non-open is a problem as the structure of $\text{graph}(f)$ is so nice that it seems like $f$ can be analytically extended to an open set containing $A$.
I've seen proofs that the converse is true, but as those proofs rely on explicitly constructing a diffeomorphism using the implicit function theorem they're not amenable to reverse engineering.