Rudin's proof of the change of variable theorem I am having trouble with Rudin's proof of the change of variable theorem for multiple integrals. The theorem is for 1-1 $\mathscr{C'}$ mappings from $R^k$ into $R^k$. In theorem 10.7 just before the change of variable theorem, he proves that if $\mathbf{F}(\mathbf{x})$ is a $\mathscr{C'}$ mapping of an open set $E\subset{R^k}$ into $R^k$ with $0\in{E}$, with $\mathbf{F}(\mathbf{0})=0$ and $\mathbf{F'}(0)$ invertible, then there is a neighborhood of $\mathbf{0}$ in which the 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 $\mathbf{G}_i(\mathbf{x})$ being a primitive $\mathscr{C'}$ mapping in some neighborhood of zero, $\mathbf{G}_i(\mathbf{0})=0$, and $\mathbf{G'}_i(0)$ is invertible, and each $B_i$ is either a flip or the identity operator. In the change of variable theorem, he claims that we can write $T(\mathbf{x})$, our 1-1 $\mathscr{C'}$ on $R^k$ mapping, as $$\mathbf{T}(\mathbf{x})=\mathbf{T}(\mathbf{a})+B_1\cdots B_{k-1}\mathbf{G}_k\circ \cdots \mathbf{G}_1(\mathbf{x-a})$$ If $\mathbf{T}(\mathbf{x})$ is linear, I understand how theorem 10.7 applies, because $\mathbf{T}(\mathbf{0})=\mathbf{0}$ for all linear transformations and we can apply the theorem to $\mathbf{T}(\mathbf{x-a})$. But if T is not linear, how does he arrive at this equation? 
Secondly, even if the equation does hold, clearly $\mathbf{T}(\mathbf{x-a})$ is composition of primitive $\mathscr{C'}$ mappings and flips, but why is $\mathbf{T}(\mathbf{x})$. Doesn't the addition of the constant term $\mathbf{T}(\mathbf{x})$ change things? 
 A: $$\mathbf{T}(\mathbf{x}) = \mathbf{T}(\mathbf{a}) + B_1\dotsb B_{k-1} \mathbf{G}_{k}\circ \dotsb \circ \mathbf{G}_1(\mathbf{x} - \mathbf{a})\tag{1}$$
represents $\mathbf{T}$ as the composition of two translations, $k$ primitive mappings, and at most $k-1$ flips. We obtain that representation by looking at the function
$$\mathbf{U}\colon \mathbf{y} \mapsto \mathbf{T}(\mathbf{a} + \mathbf{y}) - \mathbf{T}(\mathbf{a})\tag{2}$$
defined on the open set
$$E - \mathbf{a} = \{ \mathbf{x} - \mathbf{a} : \mathbf{x} \in E\}.$$
Since $\mathbf{U}(\mathbf{0}) = \mathbf{0}$, we have the representation
$$\mathbf{U}(\mathbf{y}) = B_1\dotsb B_{k-1} \mathbf{G}_{k}\circ \dotsb \circ \mathbf{G}_1(\mathbf{y})\tag{3}$$
from theorem 10.7, and since
$$\mathbf{T}(\mathbf{x}) = \mathbf{T}(\mathbf{a}) + \bigl(\mathbf{T}(\mathbf{a} + (\mathbf{x}-\mathbf{a})) - \mathbf{T}(\mathbf{a})\bigr) = \mathbf{T}(\mathbf{a}) + \mathbf{U}(\mathbf{x} - \mathbf{a}),$$
$(1)$ is a direct consequence of $(3)$.
What Rudin does not explicitly say in the proof of the change-of-variables formula is that the formula also holds for translations (a translation is, by the way, a composition of at most $k$ primitive $\mathscr{C}'$ maps, which then shows again that the transformation formula holds for translations). So $(1)$ is a representation of $\mathbf{T}$ as a composition of maps for which the transformation formula holds, and hence it also holds for $\mathbf{T}$.
