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Wikipedia offers the following definition for an (embedded) submanifold:

An embedded submanifold (also called a regular submanifold), is an immersed submanifold for which the inclusion map is a topological embedding.

I've been wondering if one could not equivalently define a submanifold like this:

$(\ast)$ Let $M$ be a smooth $m$-manifold and let $N$ be a subset. Then $N$ is an $n$-submanifold if and only if it is itself a $n$-manifold when endowed with the subspace topology.

Is this definition equivalent to the usual definition with embedding?

Edit!

It appears that "my" definition is in fact common, too.

I seem to have found it in this book on page 5:

1.5. Definition. A subset $M \subset \mathbb R^n$ is called differentiable submanifold of $\mathbb R^n$ of dimension $m \le n$ if to each $x\in M$ there corresponds an invertible germ $\widetilde{\phi}:(\mathbb R^n,x) \to (\mathbb R^n,0)$ such that $\widetilde{\phi} (M,x) = (\mathbb R^m,x)\subset (\mathbb R^n,x)$ ($\mathbb R^m$ linearly embedded in $\mathbb R^n$ for $m \le n$)

This is exactly the same as $(\ast)$ if in $(\ast)$ we let $M = \mathbb R^n$, right?

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  • $\begingroup$ Technically that is weaker even than the second condition of the inclusion map being a topological embedding...nothing says that the subspace topology can't be a different manifold topology than the original! It would also imply that the immersion condition is superfluous, which I doubt (but don't know). $\endgroup$ – Matt Samuel Sep 2 '15 at 2:53
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    $\begingroup$ The question is basically "why do we need an embedding to be an immersion?" in this definition. A pathological reason is given here: math.stackexchange.com/a/1136737/192336 Basically, we want not only $f(N)$ to be a topological manifold in the subspace topology, but we also want the tangent space to behave nicely with respect to this map. $\endgroup$ – Moya Sep 2 '15 at 2:57
  • $\begingroup$ @Moya Sounds like an answer to me! $\endgroup$ – Matt Samuel Sep 2 '15 at 2:58
  • $\begingroup$ Didn't you mean to say $N$ is a smooth $n$-manifold? $\endgroup$ – Ted Shifrin Sep 2 '15 at 6:01
  • $\begingroup$ @TedShifrin Yes I did. Does it make a difference? $\endgroup$ – a student Sep 3 '15 at 1:54

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