Is the following universal property of holomorphic functions true? Suppose that we are given a smooth $f\in C^\infty(\mathbb{\mathbb{R}^2},\mathbb{C})$ and define $\iota$ to be the smooth embedding
$$
\begin{aligned}
\iota :\mathbb{R}^2&\to\mathbb{C}^2\\
(x,y)&\mapsto (x+iy,x-iy)=((x,y),(x,-y)).
\end{aligned}
$$
Is it true that there exists a unique holomorphic $\hat{f}\in\mathscr{O}({\mathbb{C}^2},\mathbb{C})$ such that $f=\hat{f}\circ \iota$?
I suspect this universal property to be true, but my background on holomorphic functions in several variables is a bit weak and I could be wrong since I never saw it mentioned anywhere. Any hints or full answers are much appreciated.
This question has occurred to me while studying complex manifolds. The aim is to make sense of the formal expression $f(z,\overline{z})$ used in that context for a smooth $f$.
 A: No.  Because that will imply that $f$ has a power series in $z$ and $\bar{z}$ and hence in $x$ and $y$.  So you'd need $f$ to be real-analytic.  But it gets even more complicated than that.  Unlike for complex-analytic functions, if $f$ is real-analytic on $\mathbb{R}^2$ it does not mean that the series at any point will converge on all of $\mathbb{R}^2$. This is true in $\mathbb{R}$ too, think of $\frac{1}{1+x^2}$ which is real analytic in $\mathbb{R}$ but it is not a restriction of a function holomorphic in $\mathbb{C}$.  So a real-analytic function in ${\mathbb{R}}^2$ is a restriction of a function holomorphic on $U$, where $U$ is some neighborhood of $\iota(\mathbb{R}^2)$ in your notation, but $U$ does not need to equal $\mathbb{C}^2$.
It is common practice among some people in several complex variables (including me) to write $f(z,\bar{z})$ for a smooth function instead of $f(z)$.  However, in that case, the function is only defined on the "diagonal" (so $\iota(\mathbb{R}^2)$ in your notation).  So we're thinking of $f$ being defined on a (totally-real) submanifold of $\mathbb{C}^2$.  The point of that notation is because chain rule works almost as if $z$ and $\bar{z}$ were separate variables if you use the Wirtinger operators, while writing $f(z)$ almost makes it seem that $\frac{\partial}{\partial \bar{z}} f$ is zero.  However, this practice is not universal, and is not even useful in every computation.
