Suppose ,, are nonnegative integers, $≥$,$≥$,$$ is a $\mathscr b$ mapping of an open set ⊂$^{}$ into $^{}$, and $′()$ has rank $r$ for every $∈$.
Fix $∈$, put $=′()$, Let $_{1}$ be the range of $A$, and let $$ be the projection in $^{}$ whose range if $_{1}$. Let $_{2}$ be the null space of $$.
Then there are open sets $$ and $$ in $^{}$, with $∈$,$⊂$ and there is a 1-1 mapping $$ of $$ onto $$(whose inverse is also of class $\mathscr b$) such that
$$(())=+() (∈) $$ where $$ is a $\mathscr b$ mapping of open set $()⊂_{1}$ into $_{2}$.
Here is the proof:
If $r=0$, Theorem 9.19 shows that $F(x)$ is constant in a neighborhood $U$ of $a$, and the equation above holds trivially. With $V=U$,$H(x)=x$,$ \phi(0)=F(a)$
I couldn't understand that why the fact that $F(x)$ is a constant in a neighborhood $U$ of $a$ when r is equal of 0 makes this equality: $$(())=+() (∈) $$ trivial.
I also don't understand why $\varphi(0)$ is equal of $F(a)$.
Any help would be appreciated.