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The statement is: Prove any $k$-dimensional regular submanifold of a $k$-dimensional manifold is an open subset. The hint is show that any submersion is an open map.
A regular sub-manifold $S$ of $M$ is such that $\forall p \in S$, there is a chart $(U ,\phi)$, $\phi : U \rightarrow V$ such that $\phi(U \cap S) = (\mathbb{R}^{k} \times \{ 0 \}^{n-k}) \cap V$. A submersion of manifolds $f : M \rightarrow N$ is a smooth map with $D_{p}(f) : T_{p} M \rightarrow T_{f(p)}N$ surjective $\forall p \in S$.

I don't really have much intuition for either of these concepts. Assuming the hint is true. How do I show that the regular submanifold must be an open set. I am thinking that both the manifolds having dimension $k$ implies some submersion between them, and perhaps this would show that the submanifold is an open set, but I am rather confused.

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The inclusion map is a submersion, since the derivative is an isomorphism at each point, hence a surjection. (Presumably you've proved in class or as an exercise that submersions are always open maps. In this case, you could apply the inverse function theorem directly.)

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    $\begingroup$ So $f : S \rightarrow M$ is the inclusion map. I just want to know in more detail why $f$ is a submersion. What exactly does it mean for the derivative to be an isomorphism? Does it mean that for any vector $w$ in the tangent space of $f(p)$, there is a vector in the tangent space of $p$ such that $D_{p}(f)(v) = w$? Is $f$ an isomorphism because $D_{p}(f) : T_{p}(M) \rightarrow T_{f(p)}(M) = T_{p}(M)$ is the identity map? Also, where do we use the assumption that $S$ is $k$-dimension and regular? Is the inclusion map not a surjection without these assumptions, and if so why? $\endgroup$ – user100101212 Feb 14 '20 at 0:06
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    $\begingroup$ The derivative of the inclusion map $M\hookrightarrow N$ is in general an injection $T_pM\to T_pN$. Since the manifolds have the same dimension, an injective linear map becomes an isomorphism. $\endgroup$ – Ted Shifrin Feb 14 '20 at 0:14
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    $\begingroup$ So are we appealing to the fact that an injective linear map between vector spaces of equal finite dimension is an isomorphism? What is the dimension of $T_{p}(S)$, where $p \in S$. Is it $k-1$? Are we saying for fixed $p \in S$, $T_{p}(S)$ is isomorphic to $T_{p}(M)$ because the derivative is an injective linear map between vector spaces of equal dimension? This then would yield that the inclusion map is a submersion. Lastly, where do we use the assumption that $S$ is a regular submanifold (do we need it)? $\endgroup$ – user100101212 Feb 14 '20 at 1:08
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    $\begingroup$ There can't be a non-regular submanifold when they have the same dimension, but the definition allows you to check immediately that the inclusion map is a local diffeomorphism. $\endgroup$ – Ted Shifrin Feb 14 '20 at 1:55

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