Confusion with Edwards's Proof of Inverse Mapping Theorem: The convergent sequence. This question arose while working through the proof of the inverse
mapping theorem (Theorem III 3.3) in C.H. Edwards, Jr.'s Advanced
Calculus of Several Variables.
Edit to add definitions:
$f:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$ is a $\mathscr{C}^{1}$ mapping in a neighborhood of the point $a$. $T=df_{a}:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$ is the differential of $f$ at $a$.
The follow are established in the text
$g_{0}(y)=a$
$\tau_{a}(x)=x+a$
$T\circ\tau_{a}^{-1}\circ g_{k+1}(y)=T\circ\tau_{a}^{-1}\circ g_{k}(y)-[f(g_{k}(y))-y]$.
Edwards instructs the student to apply $\tau_{a}\circ T^{-1}$
to both sides of this expression. What I get is
$g_{k+1}(y)=g_{k}(y)-\tau_{a}\circ T^{-1}[f(g_{k}(y))-y]$
$g_{k+1}(y)=g_{k}(y)-T^{-1}[f(g_{k}(y))-y]-a$.
Edwards gives the generic induction step as
$g_{k+1}(y)=g_{k}(y)-T^{-1}[f(g_{k}(y))-y]$.
Now, according to my result the first induction step is
$g_{1}(y)=g_{0}(y)-T^{-1}[f(g_{0}(y))-y]-a$
$=a-T^{-1}[f(a)-y]-a$
$=-T^{-1}[f(a)-y]$
So the $-a$ is canceled. But I don't see how that would be inherited
in subsequent steps. For example, 
$g_{2}(y)=-T^{-1}[f(-T^{-1}[f(a)-y])-y]-a$
seems to be the obvious next term.
Where is the error?
 A: It's not a mistake in the book. The point is that the translation $\tau_a$ is not linear, but you treated it as if it were.
We don't have
$$\tau_a\bigl(\tau_a^{-1}(g_k(y) - [f(g_k(y))-y]\bigr) = \tau_a\bigl(\tau_a^{-1}(g_k(y))\bigr) - \tau_a[f(g_k(y))-y]$$
but
$$\tau_a\bigl(\tau_a^{-1}(g_k(y) - [f(g_k(y))-y]\bigr) = \tau_a\bigl(\tau_a^{-1}(g_k(y))\bigr) - [f(g_k(y))-y].$$
That error would be easier to avoid if instead of $T\circ \tau_a^{-1}\circ g_k(y)$ one wrote
$$T\bigl(g_k(y) - a\bigr).$$
Then it's easy to see that
$$T\bigl(g_{k+1}(y) - a\bigr) = T\bigl(g_k(y) - a\bigr) - [f(g_k(y)) - y]$$
is transformed first to
$$\bigl(g_{k+1}(y) - a\bigr) = \bigl(g_k(y) - a\bigr) - T^{-1}[f(g_k(y))-y]$$
and then to
$$g_{k+1}(y) = g_k(y) - T^{-1}[f(g_k(y))-y].$$
A: WRONG:I am convinced the book is wrong.  The terms of the series should be as I have shown them.
As Daniel Fischer so generously pointed out, I was treating $\tau_{a}(x)=x+a$ as a linear function.  Actually, my mind was treating $\circ$ as a distributive operator.
$\tau_{a}(x)=x+a$
$\tau_{a}(\xi x)+\tau_{a}(\sigma y)=\xi x+\sigma y+2a$
$\tau_{a}(\xi x+\sigma y)=\xi x+\sigma y+a\ne\tau_{a}(\xi x)+\tau_{a}(\sigma y)$
If $x_{n+1}=\phi(x_{n})$ then
$\tau_{a}(x_{n+1})=\tau_{a}(\phi(x_{n}))$
$=x_{n+1}+a=\phi(x_{n})+a$
So applying $\tau_{a}$ to both sides of the induction rule equation
has a null effect.
