# Why is the image of a smooth embedding $f: N \to M$ an embedded submanifold?

I'm reading An Introduction to Manifolds (Tu) and got confused on p.123 Theorem 11.13. Let me briefly explain what was done before that.

The author defines an embedding between two manifolds $f: N\rightarrow M$ as an injective immersion (differential is injective everywhere) with the additional property that $f(N)$ as a subspace of $M$ is homeomorphic via $f$ to $N$. Then a canonical form theorem is proven, namely, that for any point $p \in N$, there are charts $(U,\phi = x^1, \ldots, x^n) \ni p$ and $(V,\psi = y^1, \ldots, y^m) \ni f(p)$, such that the transition map takes the following form:

$$\psi \circ f \circ \phi^{-1} : (r^1,\ldots,r^n) \mapsto (r^1,\ldots,r^n,0,\ldots,0).$$

(I guess it is implicitly assumed that $f(U) \subset V$).

Now in the process of proving that $f(N) \subset M$ is an embedded submanifold, he states that $f(U) \subset V$ is defined by the vanishing of the coordinates $y^{n+1},\ldots,y^m$. I can prove that $$f(U) \subset \{q \in V: y^{n+1}(q) = \cdots = y^m(q)\}$$ but not the reverse inclusion. Can anyone explain why this should be true?

The reverse inclusion doesn't necessarily follow from the canonical form theorem you've stated, since $V$ can just happen to be too large (or, if you like, $U$ can be too small): given $U$ and $V$ as you have one can always replace $U$ by a much smaller neighborhood $U'$ of $0$, and still guarantee that $U'$ and $V$ have charts satisfying your theorem -- and also be sure that the inclusion $\supset$ can't hold.
The solution, of course, is just to prove a slightly better canonical form theorem, by reducing the possible size of $V$:
Theorem. For any $p \in N$, there exist charts $(U, \phi = x^1, \dots, x^n)$ and $(V, \psi = y^1, \dots, y^m)$ satisfying $$\psi f \phi^{-1} : (r^1, \dots, r^n) \mapsto (r^1, \dots, r^n, 0, \dots, 0)$$ and such that $$f(U) = \lbrace q \in V : y^{n+1}(q) = \cdots = y^m(q) = 0 \rbrace.$$
Proof. The proof is simple once you have your canonical form theorem; if $(U, \phi)$ and $(V', \psi)$ are the charts guaranteed by your theorem, then you can take $$V = \psi^{-1} \Big( \psi(V') \cap (\phi(U) \times \mathbb{R}^{m-n}) \Big).$$ Then $(U, \phi)$ and $(V, \psi|_V)$ satisfy the conditions of this new theorem.