Real analytic inverse and implicit functions

Complexify $f$ to $F$ on a conjugation-invariant polydisc by F1. Its complex Jacobian at the real point $a$ is the same real matrix as $Df(a)$, so its determinant is nonzero. F2 supplies a holomorphic inverse $G$ on an open neighborhood of $f(a)$. Shrink its domain to a conjugation-invariant polydisc $V$ on which both $G(w)$ and $\overline{G(w)}$ lie in the injectivity neighborhood of $F$. This is possible by continuity at $f(a)$, since both limits are $a$.

Real coefficients give $F(\bar{z})=\overline{F(z)}$. Thus $F(\overline{G(w)})=\bar{w}=F(G(\bar{w}))$ for $w\in V$. Injectivity gives $G(\bar{w})=\overline{G(w)}$, so $G$ maps the real slice into the real slice. Each component is continuous and separately holomorphic (restrict its complex differential to a coordinate line), so F3 expands it at the real centre $f(a)$. Conjugating this series and using the displayed identity and F4 shows every coefficient equals its conjugate. The expansion therefore restricts to a real analytic inverse.

Set $H(x,y)=(x,P(x,y))$. Its derivative at $(a,b)$ has block matrix $\left(\begin{array}{cc}I& 0\\ D_{x}P& D_{y}P\end{array}\right)$, whose determinant is $\det D_{y}P\mathrm{\ne }0$. Steps 1.1–2.1 give a real analytic inverse $K$ near $(a,0)$. The first coordinate identity $H(K(x,z))=(x,z)$ forces $K(x,z)=(x,k(x,z))$. Define $g(x)=k(x,0)$ after shrinking to a product neighborhood.

The identity $H(K(x,0))=(x,0)$ implies $P(x,g(x))=0$ and $g(a)=b$. Conversely, if $(x,y)$ is in the chosen inverse neighborhood and $P(x,y)=0$, then $H(x,y)=(x,0)$, hence $(x,y)=K(x,0)$ and $y=g(x)$. This proves both the graph description and local uniqueness.