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A subset $S \subseteq M$ is a regular submanifold of dimension $k$ of a smooth manifold $M$ of dimension $n$ if for every point $p \in S$, there exists a chart $(U_p, \varphi_p)$ around $p$ in the smooth structure of $M$ such that the intersection $U_p \cap S = \varphi_p'(U_p)$ where $\varphi_p'$ some function which may be obtained by sending $n - k$ of the coordinate functions of $\varphi_p$ to zero.

The above is supposed to be a definition, but I don't understand how it defines the topology or the smooth structure. Is the subspace topology implicit in the above definition? Or is it supposed to be another assumption?

Let $\varphi_S: U_p \cap S \to \mathbb{R}^k$. How can we show $(U_p \cap S, \varphi_S)$ is a chart on $S$? Does this process of constructing charts uniquely determine the smooth structure?

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    $\begingroup$ That those $\varphi_S$ forms a chart on $S$ is a direct checking. Did you check that? $\endgroup$ May 11, 2021 at 4:55
  • $\begingroup$ I don't quite see it. Are the adapted charts also smoothly compatible charts on $M$? I don't see how to prove that the charts $\varphi_S$ are smoothly compatible with each other. $\endgroup$
    – lanf
    May 11, 2021 at 14:15
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    $\begingroup$ Given any two $\varphi_p, \varphi_q$ you obtain two maps $\varphi^S_p : U_p \cap S \to \mathbb R^k$ and similar for $\varphi^S_q$. Try to write $(\varphi^S_q)^{-1} \circ \varphi^S_p$ in terms of $\varphi_q^{-1}\circ \varphi_p$. $\endgroup$ May 11, 2021 at 16:35

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Yes, $S$ is given the subspace topology inherited from $M$. But the phrase

there exists a chart $(U_p, \varphi_p)$ around $p$ in the smooth structure of $M$ such that the intersection $U_p \cap S = \varphi_p'(U_p)$ where $\varphi_p'$ is some function which may be obtained by sending $n - k$ of the coordinate functions of $\varphi_p$ to zero

does not make much sense. The chart $\varphi_p : U_p \to V_p$ is a homeomorphism onto an open $V_p \subset \mathbb R^n$. Now the set $U_p$ occurs on both sides of the equation $U_p \cap S = \varphi_p'(U_p)$ and therefore $\varphi_p'$ must be a map on $U_p$ and taking values in $M$ - but this has no coordinate functions which can take the value $0$. What is meant is that $\varphi_p(U_p \cap S) = V_p \cap \mathbb R^n_k$, where $\mathbb R^n_k = \{(x_1,\dots,x_n) \in \mathbb R^n \mid x_{k+1} = \dots = x_n = 0\}$. Then you can define $$\varphi'_p= \pi^n_k \circ \varphi_p \mid_{U_p \cap S}: U'_p = U_p \cap S \to V'_p $$ where $\pi^n_k$ denotes projection onto the first $k$ coordinates and $V'_p = \pi^ n _k(V_p \cap \mathbb R^n_k)$.

Then you can prove that the $\varphi'_p$ form an atlas on $S$ which has smooth transition functions and thus determines a smooth structure on $S$.

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