An open set in an $n$-manifold is clearly a submanifold of the same dimension as its containing manifold (see open manifolds).

Now, given an $n$-manifold $M$, is it true that a set, to be the underlying set of a submanifold of $M$ with dimension $n$, must be open?

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    $\begingroup$ This depends on whether you allow manifolds with boundary. $\endgroup$ Nov 22, 2011 at 15:00
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    $\begingroup$ The OP is clearly learning the basics of differential geometry. Let's try not to make him more confused with subtleties... $\endgroup$
    – PseudoNeo
    Nov 22, 2011 at 15:15
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    $\begingroup$ Converse: math.stackexchange.com/questions/2144470/… $\endgroup$
    – user636532
    Jul 21, 2019 at 9:19

2 Answers 2



In general, $N \subset M$ is a submanifold if you can find for every $x \in N$ an open neighbourhood $U \subset M$ of $x$, an open neighbourhood $V \subset \mathbb R^{\dim M}$ of $0$ and a diffeomorphism $\Phi : U \to V$ such that $\Phi(x) = 0$ and $\Phi(N) = (\mathbb R^{\dim N} \oplus 0) \cap V$. (It is legit to talk about diffeomorphisms because $M$ is a manifold).

Now, if $\dim N = \dim M$, the condition implies $\Phi(N) = V$. In other words, the open neighbourhood $U$ (the domain of the chart) must lie in $N$. So, $N$ contains an open neighbourhood of each of its points, and it is therefore open.

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    $\begingroup$ You mean $\Phi(U\cap N)$ instead of $\Phi(N)$, right? $\endgroup$
    – geometricK
    Mar 16, 2017 at 7:10

Let f be embedding of submanifold N into M. I only need to prove f(N) is open. Now consider Df(p) from tangent space of N at p to tangent space of M at f(p).

Now Df(p) 1-1 (by defintion) of submanifold). But dimN=dimM. Using Rank Nullity theorem we get Df(p) 1-1 onto. Hence f is local diffeomorphism at each point.

Hence f is diffeomorphism.(although proof of this also goes in same way as @PseudoNeo gave)

In particular f homeomorphism. Hence f(N) open.

  • $\begingroup$ Are you sure you are not missing out of any kind of invariance of domain? $\endgroup$
    – user636532
    Jun 21, 2019 at 13:25
  • $\begingroup$ Yes for this fact, a big theorem like Invariance of domain is not needed. $\endgroup$
    – Sushil
    Jun 21, 2019 at 14:42

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