# GP $1.2.9(d)$: $d(f\times g)_{(x,y)}=df_x\times dg_y$

Guillemin-Pollack 1.2.9(d):

Let $f: X\rightarrow X', Y\rightarrow Y'$ be smooth maps. Prove that $$d(f\times g)_{(x,y)}=df_x\times dg_y$$

(Note that in this book a manifold is a subset of $R^n$ locally diffeomorphic to an open subset of $R^k$.)

I'm not sure how to do this. Here is what

Since the differential is defined in terms of local parametrizations, start with some parametrizations. Let $U\times V$ be open in $R^k\times R^l$ and $\phi\times \psi:U\times V\rightarrow X\times Y$ be a local parametrization around $(x,y)$. Similarly let $U'\times V'$ be open in $R^m\times R^n$ and $\phi'\times \psi':U'\times V'\rightarrow X'\times Y'$ be local parametrizations around $(x',y')$. Assume for simplicity that $(0,0)\mapsto (x,y)$ and $(0,0)\mapsto (x',y')$ via the maps above. The differential is defined as follows:

$$df_{(x,y)}=d(\phi'\times \psi')_{(0,0)}\circ dh_{(0,0)}\circ d(\phi\times \psi)^{-1}_{(x,y)}$$ where $h=(\phi'\times \psi')^{-1}\circ f\circ (\phi\times \psi)$.

(Of course $(\phi\times\psi)^{-1}=\phi^{-1}\times \psi^{-1}$ and similarly for $(\phi'\times \psi')^{-1}$.)

If I apply $df_{(x,y)}$ to $(u,v)$, I first need to evaluate $d(\phi\times \psi)^{-1}_{(x,y)}(u,v)$. But I don't see how this expression can be simplified, since $\phi^{-1}\times \psi^{-1}$ is again a map on $X\times Y$, and I need to prove the result for maps like this... How should I proceed?