Showing two definitions of embedded submanifold are equivalent Let $E$ be a subset of an $m$-dimensional smooth manifold $M$.  Then $E$ is an embedded submanifold of $M$ of dimension $k$ if there is a manifold structure on $E$ such that $E$ has the induced topology from $M$, and the inclusion map is smooth and injective on tangent spaces.  I want to show that for each $p \in E$, there exists a chart $V,\psi$ of $M$ containing $p$, such that
$$\psi(V \cap E) = \psi(V) \cap \mathbb{R}^k$$
where we identify $\mathbb{R}^k$ in $\mathbb{R}^m$ via $(x_1, ... , x_k) \mapsto (x_1, ... , x_k,0,...,0)$.
My attempt: by the constant rank theorem, there exist charts $U,\phi$ and $V,\psi$ of $E$ and $M$, such that $p \in U \subseteq V$, and such that 
$$\psi \circ \phi^{-1}(x_1, ... , x_k) = (x_1, ... , x_k,0,...,0)$$
for all $(x_1, ... , x_k) \in \phi(U)$.
Shrink the open set $V$ so that $U = V \cap E$, and replace $\psi$ by its restriction to the smaller $V$.  Then obviously
$$\psi(V \cap E) \subseteq \psi(V) \cap \mathbb{R}^k$$
However, I don't think we are guaranteed equality here.  As far as I can tell, there can be a point $q \in V$, not in $E$, such that $\psi(q) = (q_1, ... , q_k,0,...,0)$.
I would appreciate any hint for how to solve this.  
 A: You have to use the hypothesis that $E \subset M$ has subspace topology. By Constant Rank Theorem we have appropriate charts centered at $p$, such that $\iota|_U : U \to V$ has representation
$$
(x^1,\cdots,x^k) \mapsto (x^1,\cdots,x^k,0,\cdots,0) .
$$
After this, we need to choose a better coordinate charts, such as small coordinate balls $U_0, V_0$ such that, $U_0 \subset U$ and $V_0 \subset V$. Note that $U_0$ still a $k$-slice in $V_0$. By hyposthesis, there are open set $H$ in $M$ s.t $U_0 = H \cap E$. I'll leave it to you to choose how shrink $V_0$ to another smaller chart $V'$ such that $V' \cap E = U_0$. This is immidiately solved because $U_0$ is a $k$-slice. 
A: Thanks to Sou for the suggestion of choosing an appropriate open $V_0 \subseteq V$.
Take $U,V,\phi,\psi$ as in my post, where $V \cap E = U$.  We have $\psi(V \cap E) = \psi(U) \subseteq \mathbb{R}^k \cap \psi(V)$, but the containment might not be proper.  However, $\psi(U)$ is open in $\mathbb{R}^k$ (if we stop identiying $\mathbb{R}^k$ as a subspace of $\mathbb{R}^m$, then $\psi(U) = \phi(U)$ with $\phi(U)$ open there).  Hence $F :=\mathbb{R}^k - \psi(U)$ is a closed set in $\mathbb{R}^m$, and so $\psi(V) - F$ is open in $\psi(V)$.  
Let $V_0 := \psi^{-1}(\psi(V) - F)$, an open set of $M$.  
Then $V_0$ contains $U$, $V_0 \cap E = U$, and we do get the desired equality
$$\psi(U) = \psi(V_0 \cap E) = \psi(V_0) \cap \mathbb{R}^k$$
