I am trying to understand Example 6.17 from Loring Tu's An Introduction to Manifolds (Second Edition, page no. 65), which is about the smoothness of a projection map. It is given below.
The solution to the example problem is a straightforward application of the definition of a smooth function between two smooth manifolds and the application of the definition is itself clear to me. What confuses me is that it is not clear how it is proved that
$$\phi \circ \pi \circ (\phi \times \psi)^{-1}: (\phi\times\psi)(U\times V) \subset \mathbb{R}^{m+n} \to \phi(U) \subset \mathbb{R}^m$$ is smooth on $(\phi\times\psi)(U\times V)$.
My Attempts
In particular, I calculate,
\begin{eqnarray}
&& \left(\phi \circ \pi \circ (\phi \times \psi)^{-1}\right) \left( (\phi \times \psi) (p, q)\right), \quad (p, q) \in (U \times V) \\ \\
&=& \left(\phi \circ \pi\right) (p, q) \\
&=& \phi \left(\pi (p, q) \right) \\
&=& \phi(p).
\end{eqnarray}
As the domain of $\phi \circ \pi \circ (\phi \times \psi)^{-1}$ and $\phi$ are different, I can not argue that as $\phi$ is smooth, $\phi \circ \pi \circ (\phi \times \psi)^{-1}$ is smooth as well at the point in their common domain. Here $\phi \circ \pi \circ (\phi \times \psi)^{-1}$ has a domain $(\phi\times\psi)(U\times V) \subset \mathbb{R}^{m+n}$ and, $\phi$ has a domain $U$.
That is I am certainly missing the argument here.
My Questions
What is that argument which proves $\phi \circ \pi \circ (\phi \times \psi)^{-1}$ to be smooth on $(\phi\times\psi)(U\times V) \subset \mathbb{R}^{m+n}$?
Best Answer
Your argument doesn't really prove much. All you said is that $(\phi\circ \pi)(p,q) = \phi(p)$. What the arugment in the book shows is that $\phi \circ \pi \circ (\phi\times \psi)^{-1}$ is a smooth map between open subsets of cartesian spaces. How? It shows that for any $(a,b) \in (\phi\times \psi)[U\times V] =:L \subset \Bbb{R}^m \times \Bbb{R}^n$, we have \begin{align} (\phi \circ \pi \circ (\psi\times\psi)^{-1})(a,b) = a \end{align} To repeat in different words, the representative of $\pi$ with respect to the charts $(U\times V, \phi\times \psi)$ on the domain and $(U,\phi)$ on the target is simply $(a,b) \mapsto a$. This is clearly the restriction of the standard projection map \begin{align} \text{pr}_{\Bbb{R}^m}: \Bbb{R}^m \times \Bbb{R}^n\to \Bbb{R}^m \end{align} to the open subset $L$ I defined above. The canonical projection map $\text{pr}_{\Bbb{R}^m}$ is clearly smooth (it is even linear, which means it is in fact also an analytic function), therefore its restriction to the open subset $L$ is also smooth (because smoothness is a local property).
So, this shows $\pi$ is smooth on the open subset $U\times V$, and in particular at the point $(p,q)$. Finally, since $(p,q)$ was taken arbitrarily, it means you can keep repeating this argument at any point of $M\times N$, and hence $\pi$ is smooth on all of $M\times N$.