Clarification on the proof for the Inverse Mapping Theorem presented by Serge Lang In the book we are presented a proof for the inverse mapping theorem which I understand for a) and b) but fail to fully understand understand for c). 


My approach:
The way I understand it is that after we let $w=z-z_0$ and define $F(w)= f(z)-f(z_0)= \sum a_nw^n$ . Then we set $w=0 =>F(w)=0$. From there we apply the previous discussion for $0$ and we arrive that $G(w)$ and $F(w)$ beeing inverses. 
But from here I am puzzled, we haven't proved that $G(w)$ is analytic and Serge Lang also included the line " Let $w_0=f(z_0)$ and let $g(w)=G(w-w_o)+z_0$" ,which is the same as $g(w)=G(w)+z_o$, and I can't figure out why this line implies the result. Furthermore, in the theorem we have the bit "Suppose that $f'(z_0) \neq 0$" which I don't see beeing used anywhere in the proof. Can anybody please clarify the proof that Lang intended?
 A: This proof is not really well written. I'll try to sketch what is going on. 
First make the reduction to $z_0 = f(z_0) = 0$. Being $f$ analytic we have that the $f$ has a power series expansion at $0$ that converges in some open neighborhood of $0$. From $f'(0) \neq 0$, we see that $f$ has a formal inverse $g$ by item (a). (Here is where he uses that the derivative does not vanish). By item (b), $g$ converges in some neighborhood of $0$ i.e. $g$ is also an analytic function.
Then he proceeds to prove that $g$ is in fact an inverse to $f$ for a good choice of neighborhoods $U_0$ and $V_0$. 
Added: Here I am adding a clarification to the reduction step. I hope this is easier to follow as the coordinate changes are written as compositions of maps.
We have two translations $T_{z_0}(z)= z+z_0$ and $T_{w_0} =w+w_0$ where $w_0=f(z_0)$. Then $$F(z) = \left(T_{w_0}^{-1} \circ f \circ T_{z_0}\right) (z) = f(z+z_0)-w_0$$ 
From this we see that $F(0) = f(z_0)-w_0 =0$. Then Let $G$ be the inverse of $F$ and define $$g(w) = \left( T_{z_0} \circ G\circ T_{w_0}^{-1}\right)(w)= G(w-w_0)+z_0$$
By construction, $f(z) = \left(T_{w_0} \circ F\circ T_{z_0}^{-1}\right)(z)$. Hence
$$f(g(w)) = \left(T_{w_0} \circ F\circ T_{z_0}^{-1}\circ T_{z_0} \circ G\circ T_{w_0}^{-1}\right)(w) = \left(T_{w_0} \circ F\circ  G\circ T_{w_0}^{-1}\right)(w) =\left(T_{w_0} \circ T_{w_0}^{-1}\right)(w) = w $$
