Question on Baby Rudin theorem 10.7 This is the definition which is usable in the proof .
https://i.stack.imgur.com/W6UoC.png .
Here is the theorem:
Suppose $F$ is a $\mathscr C'$ - mapping ( that means continuously differentiability) of an open set E $\subset R^n$ into $R^n$, $0 \in E $, $F(0) = 0$, and $F'(0)$ is invertible.
Then there is a neighborhood of $0$ in $R^n$ in which a representation: $$\mathbf{F}(\mathbf{x})=B_1\cdots B_{n-1}\mathbf{G}_n\circ \cdots \mathbf{G}_1(\mathbf{x})$$.
is valid.
with each $\mathbf{G}_i$ being a primitive $\mathscr{C'}$ mapping in some neighborhood of $0$, $\mathbf{G}_i(\mathbf{0})=0$, and $\mathbf{G'}_i(0)$ is invertible, and each $B_i$ is either a flip or the identity operator.
Here is the proof:
Put $F = F_1$. Assume $1 \leq m \leq n - 1,$ and make the following induction hypothesis ( which evidently holds for $m$= 1):
$V_m$ is a neighborhood of $0$, $F_m$ $\in$ $\mathscr C'(V_m)$, $F_m(0)$ = $0$, $F_m'(0)$ is invertible, and
$$P_{m-1}F_m(x) = P_{m-1}x (  x \in V_m).  (1)$$
by $(1)$, we  have:
$$F_m(x) = P_{m-1}x + \sum_{i=m}^n \alpha_i(x)e_i$$
where $\alpha_m,...,\alpha_n$ are real $\mathscr C'$-functions in $V_m$.
Hence
$F_m'(0)$$e_m$ = $\sum_{i=m}^n$ $(D_m\alpha_i)(0)$$e_i$. ( Mark this equality by ($\oplus$)).
Since $F_m'(0)$ is invertible, the left side of $(\oplus)$ is not $0$, and therefore there is a $k$ such that $m$ $\leq$ $k$ $\leq$ $n$ and ($D_m$$\alpha_k$)($0$) $\neq$ $0$.
Let $B_m$ be the flip (the definition of this is in the first link) that interchanges $m$ and this $k$ ( if $k = m$, $B_m$ is the identity ) and define
$G_m(x)$ = $x$ + [$\alpha_k(x)$ - $x_m$]$e_m$   ($x$ $\in$ $V_m$).
Then $G_m$ $\in$ $\mathscr C'(V_m),$ $G_m$ is primitive and $G_m'(0)$ is invertible, since ($D_m$$\alpha_k$)($0$) $\neq$ $0$.
The inverse function theorem shows therefore that there is an open set $U_m$, with $0$ $\in$ $U_m$ $\subset$ $V_m$, such that $G_m$ is a $1-1$ mapping of $U_m$ onto a neighborhood $V_{m+1}$ of $0$, in which $G_m^{-1}$ is continuously differentiable.
https://i.stack.imgur.com/m7BkJ.png  ( it's the inverse function theorem).
Define $F_{m+1}$ by
$F_{m+1}$($y$) = $B_m$$F_m$ $\circ$ $G_m^{-1}$($y$). ($y$ $\in$ $V_{m+1}$ ). ( We mean composition in $\circ$)
Then $F_{m+1}$ $\in$ $\mathscr C'(V_{m+1})$, $F_{m+1}$($0$) = $0$, and $F_{m+1}'$($0$) is invertible.
I don't understand how do we get the ($\oplus$) and I also don't understand why is $F_{m+1}(0)$ equal of $0$.
Any help would be appreciated.
 A: First, let me answer your first question.
$F_m'(0)$$e_m$ is going to be the $m$-th column of the $n\times n$ Jacobian matrix of $F_m$ at $x=0$. I think using the $e_i$ to define the function makes it hard to visualize it. Let's build up how $F_m$ looks in vector form so we can build its Jacobian.
$$F_m(x)=\begin{bmatrix}I_{m-1} & 0\\ 0
 & 0\end{bmatrix}x+\begin{bmatrix}0_{m-1\times1}\\ \alpha_m(x)\\ \vdots \\
  \alpha_n(x)\end{bmatrix}=\begin{bmatrix}x_1 \\ \vdots \\ x_{m-1} \\ \alpha_m(x) \\ \vdots \\ \alpha_n(x) \end{bmatrix}$$
Note that, in the first matrix above, I ommitted the dimensions of the $0$ block matrices because otherwise it looked hideous; you can deduce these dimensions unambiguously. In the second matrix I included that the $0$ block is $m-1 \times 1$.
Now the Jacobian matrix of $F_m$ at $x$ is:
$$\begin{bmatrix}I_{m-1} & 0\\ 0
 & 0\end{bmatrix}+\begin{bmatrix}0_{m-1\times n}\\ \nabla\alpha_m(x)^\text{T}\\ \vdots \\
  \nabla\alpha_n(x)^\text{T}\end{bmatrix}$$
Its $m$-th column is
$$F'_m(x)e_m=\begin{bmatrix}0_{n\times 1}\end{bmatrix}+\begin{bmatrix}0_{m-1\times1}\\ D_m\alpha_m(x)\\ \vdots \\
  D_m\alpha_n(x)\end{bmatrix}=\sum_{i=m}^{n} D_m\alpha_i(x)e_i$$
Then $F'_m(0)e_m=\sum_{i=m}^{n} D_m\alpha_i(0)e_i$.
Now the second question. In the beginning statements we had $F_m(0)=0$. Looking at the vector expression for $F_m$ above, this means $\alpha_i(0)=0$ for all $i=m,\ldots,n$, including $i=k$. Now look at the definition for $G_m$ and you'll see $G_m(0)=0$, so $G_m^{-1}(0)=0$. Finally,
$$F_{m+1}(0)=B_m F_m(G_m^{-1}(0))=B_m F_m(0)= B_m\ 0=0.$$
