I am reading a text in introductory differential geometry and it says " If $M$ is a submanifold of $\mathbb{R}^n$, and $N$ is open relative to $M$, then it follows easily from the definitions (of a submanifold) that $N$ is a submanifold of $\mathbb{R}^n$, with same dimension as $M$.

I don't see how it follows so easily. I think that since $M$ is a submanifold of $\mathbb{R}^n$, then there exists a smooth $F$ such that $F:U \subset \mathbb{R}^n \rightarrow \mathbb{R}^k$, where $k = n-m$. In other words, $F^{-1}(c) = U \cap M$. If we were to consider the restriction of $F$ to the set $U \cap M$, then by definition of relativity open (i.e. $\exists O$ open such that $O \cap M$ is relatively open), the map $F$ restricted to $U \cap M$ maps to $\mathbb{R}^k$, and the image is contained in a neighborhood of $c$. This restriction is also smooth, so the fact that a relatively open subset is a submanifold follows.

Is this all that I need? I feel I am getting mixed up somewhere.


Consider a smooth atlas on $M$, say ($\varphi_{\alpha}, U_{\alpha})$, where every $U_{\alpha}$ is open. Now for any $\alpha$ such that $U_{\alpha} \cap N \neq \emptyset$ consider $\overline{\varphi_{\alpha}}: U_{\alpha} \cap N \rightarrow \mathbb{R}^k$ defined by taking the restriction of $\varphi_{\alpha}$ over $U_{\alpha} \cap N$.

The new atlas ($\overline{\varphi_{\alpha}}, U_{\alpha} \cap N$) is a smooth atlas on $N$ (the restriction of smooth maps to open sets is smooth, and ($U_{\alpha} \cap N$)$_{\alpha}$ is a partition of $N$).

  • $\begingroup$ I see, that's what I figured but wasn't quite sure how to say it.. thanks. $\endgroup$ – LordVader007 Feb 14 at 8:12

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