Exercise 10.3

(a) (Much of this solution just repeats the proof of Theorem 10.7.) Assume $1\le m\le n-1$, and make the following induction hypothesis (which evidently holds for $m=1$):

$V_m$is a neighborhood of $\check{0}$, ${\check{F}}_m\in C^{\prime }(V_m)$, ${\check{F}}_m(\check{0})=\check{0}$, ${\check{F}}_m^{\prime }(\check{0})=I$, and for $\check{x}\in V_m$,

By $\text{(∗)}$, we have

where ${\alpha }_m,\dots ,{\alpha }_n$ are real $C^{\prime }$-functions in $V_m$. Hence, for $j=m,\dots ,n$,

Since ${ě}_m,\dots {ě}_n$ are independent, we must have

Define, for $\check{x}\in V_m$,

Then ${Ǧ}_m\in C^{\prime }(V_m)$, ${Ǧ}_m$ is primitive, and ${Ǧ}_m^{\prime }(\check{0})=I$ by $\text{(∗∗)}$. The inverse function theorem shows therefore that there is an open set $U_m$, with $\check{0}\in U_m\subset V_m$, such that ${Ǧ}_m$ is a 1-1 mapping of $U_m$ onto a neighborhood $V_{m+1}$ of $\check{0}$, in which ${Ǧ}_m^{-1}$ is continuously differentiable, and

Define ${\check{F}}_{m+1}(\check{y})$, for $\check{y}\in V_{m+1}$, by

Then ${\check{F}}_{m+1}\in C^{\prime }(V_{m+1})$, ${\check{F}}_{m+1}(\check{0})=\check{0}$, and ${\check{F}}_{m+1}^{\prime }(\check{0})=I$ by the chain rule. Also, for $\check{x}\in U_m$,

so that, for $\check{y}\in V_{m+1}$, $P_m{\check{F}}_{m+1}(\check{y})=P_m\check{y}$. Our induction hypothesis holds therefore with $m+1$ in place of $m$.

Note that, for $\check{y}={Ǧ}_m(\check{x})$, we have

If we apply this with $m=1,\dots ,n-1$, we successively obtain

in some neighborhood of $\check{0}$. By $\text{(∗)}$, ${\check{F}}_n$ is primitive, so we can let ${Ǧ}_n={\check{F}}_n$.

(b) Let $\check{F}$ be the mapping $(x,y)\to (y,x)$ and suppose $\check{F}={Ǧ}_2\circ {Ǧ}_1$ in some neighborhood of the origin, where

are primitive mappings. Then we would have

so that

which is impossible. Trying $\check{F}={Ǧ}_1\circ {Ǧ}_2$ leads to a similar contradiction.