Holomorphic inverse function theorem and local-degree criterion

For the equivalence between claims 1 and 2, [L3] gives $f(z)-f(a)=(z-a{)}^{m}q(z)$ with $m={\mathrm{deg}}_{a}f$ and $q(a)\mathrm{\ne }0$. If $m=1$, differentiation at $a$ gives $f^{\mathrm{\prime }}(a)=q(a)\mathrm{\ne }0$; if $m>1$, it gives $f^{\mathrm{\prime }}(a)=0$. Thus $f^{\mathrm{\prime }}(a)\mathrm{\ne }0$ exactly when ${\mathrm{deg}}_{a}f=1$.

For the implication from claim 1 to claim 4, [L1] directly makes $f$ biholomorphic between neighbourhoods of $a$ and $f(a)$.

For the implication from claim 4 to claim 3, a biholomorphic restriction is bijective and hence injective on its source neighbourhood.

For the converse implication from claim 3 back to claim 2, suppose $f$ is injective on a neighbourhood $N$ of $a$. If $m={\mathrm{deg}}_{a}f>1$, take $V\subseteq N$ and $\rho >0$ from [L2] and put $w:=f(a)+{\rho }^m\mathrm{/}2$. Then $w$ has $m>1$ distinct preimages in $V$, contradicting injectivity on $N$. Since $m$ is positive, $m=1$, and step 1.1 then gives $f^{\mathrm{\prime }}(a)\mathrm{\ne }0$.

For the derivative formula, let $g$ be the inverse supplied in step 1.2. Differentiating $g(f(z))=z$ gives $g^{\mathrm{\prime }}(f(z))f^{\mathrm{\prime }}(z)=1$, so, writing $w=f(z)$, one obtains $g^{\mathrm{\prime }}(w)=1\mathrm{/}{f}^{\mathrm{\prime }}(g(w))$.