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I needed help to prove the following:

Let $k, n, m$ be elements of the natural numbers and $g : R^m \to R^n$. Prove that the graph of $g$ is an $m$-manifold of class $C^k$ if and only if $g$ is of class $C^k$

from munkres

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    Yes, and k-time continuous, right?2012-03-09

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Let $\psi\colon\mathbb R^m\times\mathbb R^n\to\mathbb R^n$ defined by $\psi(x,y):=y-g(x)$. Then $\psi$ is of class $C^k$ if and only if $g$ is of class $C^k$. We have $D\psi=\pmatrix{-Dg& I_n}$ so its rank is $n$, hence maximal. So $\psi^{-1}(\{0_n\})$ defines a sub-manifold of $\mathbb R^n\times \mathbb R^n$ of dimension $m+n-n=m$ and class $C^k$. What remains is to show that it cannot be a manifold of class $C^{k+1}$ if $g$ is not $C^{k+1}$. To see that, take a point on which this condition fails and show that there is no submersion which can do the work.