# If $f:U\to\mathbb{R}^N$ is a $C^k$-submersion, $g:f(U)\to \mathbb{R}^M$ and $g\circ f :U\to\mathbb{R}^M$ is $C^k$ then $g$ is $C^k$

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 $$C^k$$.

• Take $h\circ F^{-1}$ that is? – weymar andres Jun 15 at 16:31
• Please somebody can help me? – weymar andres Jun 16 at 13:56
• Why is he deprived of your bounty? You have accepted the answer. – 0-th User Jun 26 at 7:36
• I thought i did, is my first time, excuse me. Thanks – weymar andres Jun 26 at 13:53
• Great, no problem. Thanks. – 0-th User Jun 26 at 14:12

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.