Given $\mathcal{M} = \mathcal{M_1} \times \dots \times \mathcal{M_k}$ where $\mathcal{M_i}$ are smooth manifolds and $\phi_i:\mathcal{M_i} \to \mathcal{M_i}$ diffeomorphisms for each $i$.
Is $\phi: \mathcal{M} \to \mathcal{M}$ where $\phi = (\phi_1 , \dots , \phi_k) $ a diffeomorphism?
Here is my answer:
$\phi$ has derivative given by $\phi' : T\mathcal{M_1} \times \dots \times T\mathcal{M_k}$ where
$$ \phi' = (\phi'_1,\dots , \phi'_k) $$
as each $\phi_i$ is diffeomorphic each $\phi'_i$ is continuous. Thus by the Characteristic Property of Product Topology $\phi'$ is also continuous.
The inverse is given by $\phi^{-1} = (\phi_1^{-1} , \dots , \phi_k^{-1})$. A similiar argument shows $\phi'^{-1}$ is continuous.
Thus $\phi$ is a differentiable function with differentiable inverse. So a diffeomorphism.