If $\\f:\ U\to\ R^N$ is a submersion of class $C^k$ and $g:f(U)\to\ R^M$ is such that $g\circ f : U\to\ R^M$ is $C^k$ then $g$is $C^k$

In my attempt I know that $D_{f(p_0)}$ is onto, $p_0 \in U$ and jacobian matrix $N\times(N+p)$, $J_{f(p_0)}$ where $N$ columns are linearly independent in some order, take $T:\{1,\dots,N\} \to\ \{1,\dots,N+p\}$ injective such that the first $N$ columns $C_{T_{1}},\ldots,C_{T_{N}}$ are linearly independent and let $L:R^{N+p}\to\ R^{N+p}$ a linear isomorphism such that $L(e_i)=C_{T_i}$ for $i:1,\ldots,N$ the rest element in any place, where $e_i$ are basis for $R^{N+p}$. Take $q_0$ the only element such that $L(q_0)=p_0$ and $B=L^{-1}(A)$ where $A$ is neightborhood of $p_0$ and $h=f\circ L: B\to\ R^{N}$ satisfy that is $C^k$ and $D_h(q_0)= D_f(p_0)L$ is onto and its first columns of jacobian matrix linearly independent are situated in the first place. Let $F:A=L(B)\subset R^{N+p}\to\ R^{N+p}$ such that $F(x,y)=(f(x,y),y)$ . Then $F$ is differentiable and $\det (D_F(p_0))\neq 0$ and we can apply inverse theorem, hence exist open sets $p_0\in V$ and $F(p_0)\in W\subset R^{N+p}$ such that $F:V\to\ W$ is a diffeomorphism but $F( x,0)=(f(x,0),0)$ and $g\circ F(x,0)=h(x,0) $ is $C^k$, then $g=h\circ F^{-1}$ I am not sure about this,because is not like for all elements on the domain of $g$. I will appreciate any help or hints or other method for solving the problem, please and thanks.

  • $\begingroup$ Thank you Bernard $\endgroup$ – weymar andres Jun 16 at 19:12

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