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I am actually in the resolution of the problem Show that $\Delta$ is diffeomorphic to X, so $\Delta$ is a manifold if $X$ is. - "Differential topology" of Guillemin and Pollack (my own question), and I was wondering if a subset of a smooth manifold is itself smooth manifold (submanifold).

Assume $X$ is a manifold and we have a smooth structure on the product manifodld $X×X$ , does that make $Δ$ a smooth submanifold? I already know that the answer is simply no, but is there sufficient conditions that we would provide an affirmation to this question?

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  • Clearly a subset is not a submanifold: the square is not a submanifold of the plane (it has corners). – Alex M. May 06 '15 at 19:21
  • I already know all this, but certain subsets of a smooth manifold is itself smooth manifold (submanifold). For example, we already know that $ S^1 \subset \mathbb {R}^2 $ is a smooth manifold of $1-dimension$, and $ H= {(x, y) \in \mathbb{R^2} : x^2 + y^2 = 1~and~y>0 }$ is a smooth manifold of $1-dimension$; although $ H \subset S^1 $. The question is when does a subset of smooth manifold is itself a smooth manifold. –  May 06 '15 at 19:45

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I found the answer at my question. ''Open subsets of a smooth manifold are smooth manifolds. ''

reference : http://www-personal.umich.edu/~wangzuoq/635W12/Notes/Lec%2002.pdf