# Image of regular submanifold is a regular/embedded submanifold.

Let $$F:N\rightarrow M$$ be a smooth embedding. Then, $$F(N)$$ is an embedded submanifold of $$M$$.

So what I have tried is the following: Let $$F(p)\in F(N)$$ for some $$p\in N$$. As $$F$$ is an immersion, by the immersion theorem, there exists charts $$(U,\phi)=(U,x^1,....,x^n)$$ about $$p$$ and $$(V,\psi)=(V,y^1,.....,y^m)$$ about $$F(p)$$ on which $$F(U)\subseteq V$$ and $$(\psi \circ F\circ \phi^{-1})(x^1,.....,x^n)=(x^1,.....,x^n,0,.....,0)$$ on $$\phi(U)$$.

As $$F(N)$$ is homeomorphic to $$N$$, there exists an open set $$W$$ in $$M$$ such that $$F(U)=V'\cap F(N)$$. Hence, $$V\cap V'\cap F(N)= V\cap F(U)=F(U)$$.

Claim: $$\psi(V\cap V'\cap F(N))$$ = $$\psi(V\cap V')\cap (\mathbb{R}^{n}$$ $$\times \{0\} )$$

Now clearly, $$\subseteq$$ holds. However, I don't think the reverse inclusion holds. I mean, for $$\psi(q)\in \psi(V\cap V') \cap (\mathbb{R}^n \times \{0\})$$ , there is no immediate reason for $$q$$ to be in $$F(N)$$ as well.

To remedy this situation, I suppose we would need to construct a new chart $$W$$ such that $$W\subseteq V\cap V'$$ and $$W\subseteq F(N)$$. I would like a nudge in the right direction.

• Improve your question: state that you want to prove a statement, then give the statement (as a proposition for instance) then recall the definition of smooth embedding and embedded submanifold, then only give your attempts. – Arnaud Oct 7 '20 at 16:38

The inclusion $$\psi(V\cap V'\cap F(N))\subseteq\psi(V\cap V')\cap (\mathbb{R}^{n}\times \{0\} )$$ could indeed be strict, but note that if dim$$(N)$$<dim$$(M)$$ restricting the chart $$(V,\psi)$$ to some open set $$W$$ with $$W\subseteq F(N)$$ cannot work for dimensional reasons.
$$W=V\cap V'\cap \psi^{-1}(\phi(U)\times\mathbb R^{m-n})$$
This ensures that in addition the ''$$\mathbb R^n$$-part" in the image of $$\psi$$ only comes from points in the domain of $$\phi$$. More formally, as an injection $$\psi$$ respects intersections, so $$\psi(W)\cap(\mathbb R^n\times\{0\}) =\psi(V\cap V')\cap(\phi(U)\times\mathbb R^{m-n})\cap (\mathbb R^n\times\{0\}) =\psi(V\cap V')\cap(\phi(U)\times\{0\}) =\psi(V\cap V')\cap\psi(F(U)) =\psi(V\cap V'\cap F(N))$$ and intersecting both sides with $$\psi(W)$$ yields
$$\psi(W)\cap(\mathbb R^n\times\{0\})=\psi(W\cap F(N))$$