Let $N$ be a smooth manifold. One possible definition (I believe) for an embedded submanifold of $N$ is some $M \subset N$ that is a (smooth) manifold such that the inclusion $i : M \hookrightarrow N$ is an embedding. Another more classical definition says that for each $p \in M$ we can find a chart $\psi : V \subset N \rightarrow \widetilde{V} \subset \mathbb{R}^{n}$, let's say with $\psi(p) = 0$, such that $\psi(V \cap M) = \{(x_1, \ldots, x_n) \in \widetilde{V}\ |\ x_{m + 1} = \ldots = x_n = 0\}$. I would like to prove the first definition implies the second.
It almost works out using the local immersion form. I can find charts $\phi : U \subset M \rightarrow \widetilde{U} \subset \mathbb{R}^{m}$ and $\psi : V \subset N \rightarrow \widetilde{V} \subset \mathbb{R}^{n}$ with $\phi(p) = 0$, $\psi(p) = 0$, $U \subset V$ and $\psi \circ \phi^{-1}$ having the form $(x_1, \ldots, x_m) \mapsto (x_1, \ldots, x_m, 0, \ldots, 0)$. However, for the equality $\psi(V \cap M) = \{(x_1, \ldots, x_n) \in \widetilde{V}\ |\ x_{m + 1} = \ldots = x_n = 0\}$ to be not a mere inclusion of the left side into the right side, I believe I would need to have $U = V \cap M$. I have been trying to argue to no avail that I can always pick the charts $\phi, \psi$ satisfying this condition. Can anyone shed any light in the issue?
Best Answer
Ah, it was quite dumb it seems. First of all, we restrict $\widetilde{V}$ to $\widetilde{V} \cap O$ where $\widetilde{U} = \mathbb{R}^{m} \cap O$ and $O \subset \mathbb{R}^{n}$ is open. Now since $U$ is open in $M$, and $M$ has the subspace topology of $N$, we have $U = W \cap M$ for some $W \subset N$ open. Now replace $V$ with $V \cap W$ and replace $U$ with $V \cap W \cap M$. Now indeed $U = V \cap M$.