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Suppose that $W$ is a linear subspace of $\mathbb R^n$. Does it trivially follow that $W$ is a regular submanifold of $\mathbb R^n$?

I would take open balls and the identity function restricted to them composed with a suitable rotation as the charts satisfying the regular submanifold property. Is it correct?

(Definition: Suppose that $M$ is a smooth manifold of dimension $m$, and $N$ is a topological subspace. $N$ is called a smooth regular submanifold of $M$ provided that for each $x\in N$ there exists a chart $(U,\psi)$ in the maximal atlas for $M$ with $x\in U$ such that $\psi(U\cap N)=\psi(U) \cap (\mathbb R^n \times {0_{\mathbb R^{m-n}}})$ for some $n\le m$.)

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    $\begingroup$ yes, that works $\endgroup$ – Glougloubarbaki Jun 8 '18 at 9:07
  • $\begingroup$ Thanks @Glougloubarbaki $\endgroup$ – user555729 Jun 8 '18 at 9:10
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    $\begingroup$ By starting with a basis of $W$ and completing it to a basis of ${\Bbb R}^n$ you reduce the question to $W$ defined by $x_{n-d}=\cdots=x_n=0$ where $d={\rm dim}(W)$. $\endgroup$ – Andrea Mori Jun 8 '18 at 9:25

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