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I want to solve a system of equation \begin{align*} F_1(x_1,...,x_N; u_1,...,u_M)&=0\\ \vdots\\ F_M(x_1,...,x_N; u_1,...,u_M)&=0 \end{align*} 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 $$\varphi_i(x_1,...,x_N)=u_i,\quad 1\leq i\leq N,$$ in the form $$x_i=\psi_i(u_1,...,u_N),\quad 1\leq i\leq N.$$


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 ?

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