Last step in the proof of $ G(F):=\{(p, F(p); p \in M\}$ is an embedded submanifold of $M \times N$ of dimension $m$, $F:M\rightarrow N$ smooth

Question

Consider the definition of embedded submanifold as follows :Let $M$ be a smooth manifold of dimension $m$ and $S\subseteq M$. $S$ is an embedded manofold of dimension $k$ if for every $p \in S$, there is a chart $(U,\varphi)$ of $M$ containing $p$. s.t

$\varphi(U \cap S)=(\Bbb R^k \times \{0\})\cap \varphi(U)$

Now the problem I am solving is

Let $M$ of dim=m and $N$ of dim=n be smooth manifolds and $F:M\rightarrow N$ a smooth map Prove that the graph of F: $ G(F):=\{(p, F(p); p \in M\}$ is an embedded submanifold of $M \times N$ of dimension $m$

To prove this I followed this steps:

1. Let $p \in M$ and $(V,\psi)$ a chart of $N$ around $F(p)$.Prove that there is a chart $(U,\varphi)$ so that $F(U)\subset V$
2. Define the following map: $h:\varphi(U)\times \Bbb R^n\rightarrow \varphi(U)\times \Bbb R^n$ as $h(x,y):=(x,y-\psi\circ F\circ \varphi^{-1}(x))$, for $x \in \varphi(U), y \in\Bbb R^n $ Prove that there are open neighborhoods $\tilde U$ of $p$ and $\tilde V$ of F(p) s.t $\tilde U \subseteq U$, $\tilde V \subseteq V$ and $h|_{\varphi(\tilde U)\times \psi(\tilde V)}$ is an diffeomeorphism onto it's image and that this image is open in $\Bbb R^m\times \Bbb R^n$
3. Define $\theta:\tilde U \times \tilde V\rightarrow \Bbb R^m\times \Bbb R^n$ as $\theta(q,r):=h(\varphi(q),\psi(r))$, for $q \in \tilde U$, $r \in \tilde V$. Prove that $(\tilde U\times \tilde V, \theta)$ is a chart of $M\times N$
4. Prove that G is an embedded submanifold of $M \times N$ of dimension $m$

I already proved 1,2,3. 2 followed from the inverse function theorem

I have a doubt in the last step.

Following the definition of submanifold I already have from point 3 that $(\tilde U\times \tilde V, \theta)$ is a chart of $M\times N$, around $(p,F(p))$ It suffices to prove that

$\theta((\tilde U\times \tilde V) \cap G)=(\Bbb R^m \times \{0\})\cap \theta(\tilde U\times \tilde V)\tag{*}$.

Since $\theta$ is in particular a bijection I can distribute $\theta$ over the intersection:

$\theta((\tilde U\times \tilde V) \cap G)=\theta(\tilde U\times \tilde V) \cap \theta(G)$

and take $(q,r) \in G$ then $r=F(q)$ and $\theta((q,r))=h(\varphi(q),\psi(F(q)))=(\varphi(q),\psi(F(q))-\psi\circ F\circ \varphi^{-1}(\varphi(q)))=(\varphi(q),0)$

so $\theta(G)\subseteq \Bbb R^m \times \{0\}$. But I would like to get $\theta(G)= \Bbb R^m \times \{0\}$ so that $(*) $ is verified. How do I conclude that?

Answer 1

Since $\theta$ is in particular a bijection I can distribute $\theta$ over the intersection:
$\theta((\tilde U\times \tilde V) \cap G)=\theta(\tilde U\times \tilde V) \cap \theta(G)$

No, you cannot. $\theta$ is only defined on the subset $\tilde U\times \tilde V$ of $ M \times N$ and in general $G$ is not contained in $\tilde U\times \tilde V$. The minimal requirement for this would be $\tilde U = M$, i.e. that $M$ is covered by a single chart. And that is possible only when $M$ is diffeomorphic to an open subset of $\mathbb R^m$.

Fortunately you do not need to "distribute $\theta$ over the intersection".

You have correctly shown that $\theta((\tilde U\times \tilde V) \cap G) \subset \mathbb R^m \times \{0\}$. Thus you get the inlusion "$\subset$" in (*).

But we can prove (*) in one go. Actually we have for all $(q,r) \in \tilde U\times \tilde V$

$$\theta(q,r) = h(\varphi(q),\psi(r)) = (\varphi(q),\psi(r) - \psi(F(\varphi^{-1}(\varphi(q))))) = (\varphi(q),\psi(r) - \psi(F(q))) .$$

But $$\psi(r) - \psi(F(q)) = 0 \Longleftrightarrow \psi(r) = \psi(F(q)) \Longleftrightarrow r = F(q)$$ since the chart $\psi$ is bijective. This shows that for $(q,r) \in \tilde U\times \tilde V$

$$\theta(q,r) \in \Bbb R^m \times \{0\} \Longleftrightarrow (q,r) \in G .$$ This is nothing else than (*).