Vtyjkyuk

I want to solve a system of equation
beginalign*
F_1(x_1,...,x_N; u_1,...,u_M)&=0\
vdots\
F_M(x_1,...,x_N; u_1,...,u_M)&=0
endalign*
where $F_i$ are $M$ function with $N+M$ variables and the desired solution is of the form $$u_i=f_i(x_1,...,x_N),quad i=1,...,M.$$

Until here it's fine and it looks as a problem of implicite function. It's the next part that I don't understand.

A very important case is in the particular case where $M=N$ and $F(x;u)$ is of the form $varphi(x)-u$ where I denote $x=(x_1,...,x_N)$, $F=(F_1,...,F_M)$ and $u=(u_1,...,u_M)$. And then, it's written : we are looking to solve $$varphi(x)=u,$$ in the form of $x=psi(u)$ so that $$varphi(psi(u))=u.$$

More explicitly, if $varphi=(varphi_1,...,varphi_N)$ we want to solve the system $$varphi_i(x_1,...,x_N)=u_i,quad 1leq ileq N,$$
in the form $$x_i=psi_i(u_1,...,u_N),quad 1leq ileq N.$$

Question : In the first part we where looking for $u$ in the form of $u=f(x)$ s.t. $0=F(x,u)=F(x,f(x))$, and they say in the second part that if $$F(x,u)=varphi(x)-u,tag*$$ then we want to find $x$ in the form of $x=psi(u)$ that solve $u=varphi(x)$. I just don't get the thing. For $F(x;u)=varphi(x)-u=0$, don't we simply have that $u=varphi(x)$ solve the problem and thus $$0=F(x,u)=F(x,varphi(x)).$$
Then we are done ! What's the matter with that ?