Showing that a cartesian product of parameterizations is still a parameterization

Question

I'm working through Guilleman and Pollack's Differential Topology. Here's the problem: Let $X,Y$ be manifolds of $\mathbb R^N, \mathbb R^M$ respectively. Then, there exists $\phi: W \rightarrow X \>\>(W \subset \mathbb R^k$, $\dim X = k$) that is a local parameterization around $x \in X$ for some $x$. Similarly, for $\dim Y = l, y \in Y$, there exists some $U \subset \mathbb R^l$ and $\psi: U \rightarrow Y$ around $y$. Then define the cartesian product of these maps: $$\phi\times \psi(w,u) = (\phi(w),\psi(u))$$ The book then asks the reader to verify that this is indeed a local parameterization of $X \times Y$ around $(x,y)$.

Here's how I am approaching it: it is obvious that $\phi \times \psi$ is surjective and injective — this follows from $\phi,\psi$ being local parameterizations. Smoothness follows in this way as well.

I'm having trouble showing that $(\phi \times \psi)^{-1}$ is a smooth map. The main concern being that $X \times Y$ might not be open in $\mathbb R^{M+N}$. This problem doesn't arise for $W \times U$ because this is open in $\mathbb R^k \times \mathbb R^l$. Any help would be appreciated.

Answer 1

I prefer to use charts to see this, but they are just the inverses of the parametrizations, so there should not be any confusion. The point is that the (slice) charts $(U,\varphi)$ and $(V,\psi)$ centered $p\in X\subseteq R^N$ and $q\in Y\subseteq R^M$ respectively, combine naturally to give a chart for $X\times Y\subseteq R^{N+M}$. That is, $(U\times V,\varphi\times \psi)$ is a chart centered at $(p,q)\in R^N\times R^M\cong R^{N+M}$ and $\varphi\times \psi$ is smooth simply because the coordinate functions are.

Or you can avoid using charts altogether by noting that the inclusions $i_X:X\to \mathbb R^N$ and $i_Y:Y\to \mathbb R^M$ are smooth immersions, so all you need to check is that $i_X\times i_Y$ is, too. So, take an arbitrary $v\in T_{(p,q)}X\times Y\cong T_pX\oplus T_qY$ and compute $d(i_X\times i_Y)v=(di_Xv_1,di_Yv_2)=0\Leftrightarrow di_Xv_1=0$ and $di_Yv_2=0$ which implies that $v_1=v_2=0.$

Answer 2

If you read Guillemin and Pollack carefully (which you need to do to learn all the definitions), when $X\subset\Bbb R^n$ they define a map $f: X\to\Bbb R^\ell$ to be smooth if at each point of $f$ there is a local extension of $f$ to a smooth map on an open subset of $\Bbb R^n$.

So, if $(p,q)\in X\times Y$, you know that $\phi^{-1}$ has a local smooth extension, $\Phi$, on an open subset $U\subset\Bbb R^N$ and $\psi^{-1}$ has a local smooth extension, $\Psi$, on an open subset $V\subset\Bbb R^M$. Now you just check that the smooth map $\Phi\times\Psi\colon U\times V\to \Bbb R^k\times\Bbb R^\ell$ is in fact an extension of $(\phi\times\psi)^{-1}$.