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If $f:\ U \to \mathbb{R}^N$ is a submersion of class $C^k$ and $g:f(U)\to \mathbb{R}^M$ is such that $g\circ f : U\to\ \mathbb{R}^M$ 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 $L:\mathbb{R}^{N+p}\to\ \mathbb{R}^{N+p}$ one isomorphism linear 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 $\mathbb{R}^{N+p}$. Take $q_0$ the only element such that $L(q_0)=p_0$ and $B=L^{-1}(A)$ where $A$ is neighborhood of $p_0$ and $h=f\circ L: B\to\ \mathbb{R}^{N}$ 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 $F:A=L(B)\subset \mathbb{R}^{N+p}\to\ \mathbb{R}^{N+p}$ 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 $F(p_0)\in W\subset \mathbb{R}^{N+p}$ 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 $C^k$.

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  • $\begingroup$ Take $h\circ F^{-1}$ that is? $\endgroup$ – weymar andres Jun 15 at 16:31
  • $\begingroup$ Please somebody can help me? $\endgroup$ – weymar andres Jun 16 at 13:56
  • $\begingroup$ Why is he deprived of your bounty? You have accepted the answer. $\endgroup$ – 0-th User Jun 26 at 7:36
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    $\begingroup$ I thought i did, is my first time, excuse me. Thanks $\endgroup$ – weymar andres Jun 26 at 13:53
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    $\begingroup$ Great, no problem. Thanks. $\endgroup$ – 0-th User Jun 26 at 14:12
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Problem like these lie in the realm of differential topology, and therefore are best viewed with a differential topological viewpoint rather than an analytical viewpoint. In fact the result holds for general $C^k$ and $C^\infty$ manifolds, the proofs more or less being the same. I'll try to break down the answer given here.

If $f$ is a diffeomorphism, we are done. If we can find a submanifold $K \subseteq U$ on which $f$ is a diffeomorphism, we are also done. It's always good to keep in mind the canonical submersion and immersion, namely $$f: (x_1, \dots, x_n, \dots x_{n + l}) \mapsto (x_1, \dots, x_n), \qquad h: (x_1, \dots, x_n) \mapsto (x_1, \dots, x_n, 0, \dots, 0)$$ Obviously if $g \circ f : (x_1, \dots, x_n, \dots, x_{n + k}) \mapsto g(x_1, \dots, x_n)$ is $C^k$, then so is $g: (x_1, \dots, x_n) \mapsto g(x_1, \dots, x_n)$. To be pedantic, but ultimately illustrative, it's because we can write $g = g \circ f \circ h$, where $h$ is clearly smooth.

We can adopt this perspective by the submersion theorem, which is essentially a variant of the inverse and implicit function theorems. This says that every submersion can be viewed in a local coordinate system as the canonical submersion. If you aren't familiar with coordinates, think of them as functions on your space which captures the Euclidean ($\mathbb R^n)$ structure, e.g. $(x, y) : \mathbb C \to \mathbb R^2$ given by $$x(z) = \Re (z), \qquad y(z) = \Im (z)$$ or polar coordinates $(r, \theta) : \mathbb C \setminus [0, \infty) \to \mathbb R^2$ given by $$r(z) = |z|, \qquad \theta(z) = \operatorname{arg} z.$$ But I digress. Since $f$ is a $C^k$ submersion, $f^{-1} (p)$ is a $C^k$ manifold of dimension $l$ (another corollary of the implicit function theorem). In fact, given the right local coordinate system $x$ on $M$, we can view $f^{-1} (p)$ as a level set $$f^{-1} (p) \cap V = \{ x_{1} = \dots = x_{n} = 0 \}.$$ Notice the connection to canonical immersion $h$. To finish off, since $g \circ f$ is $C^k$ on $U$, it is also $C^k$ on the submanifold $C = \{x_{n + 1} = \dots = x_{n + l} = 0\}$ of dimension $n$, and moreover $f$ is a submersion on $C$ (check this! The point is that $C$ is transverse to $f^{-1} (p)$). Arguing by rank nullity, $f: C \to \mathbb R^n$ is a $C^k$ diffeomorphism, so it maps onto an open neighborhood $V$ of $p$ and admits a $C^k$ inverse $h$, so we write $g = g \circ f \circ h$ locally on $V$. Composition of $C^k$ functions is $C^k$, so we are done.

The gist is that $g \circ f$ is a map on $\mathbb R^{n + l}$ but we can "throw away" $l$-many coordinate directions. Think of $f$ is a submersion as saying the domain is "too big", so by removing $l$-coordinate directions and restricting ourselves to a dimension $n$ subset of $\mathbb R^{n + l}$, we can view $f$ on this submanifold as a diffeomorphism.

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  • $\begingroup$ Thank you so much $\endgroup$ – weymar andres Jun 26 at 14:11

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