product of two diffeomorphic maps are diffeomorphic? Suppose I have a diffeomorphism  $f_i: U_i \subseteq \mathbb{R}^{n_i} \to f_i(U_i) \subseteq \mathbb{R}^{m_i}$ where $U_i$ is an open subset. Does it then follow that
$$
f_1 \times f_2: U_1 \times U_2 \to f_1(U_1) \times f_2(U_2)
$$
is still diffeomorphic? I can see that the map is still bijective, but I was struggling to show that the map and its inverse are continuous. Any explanation would be appreciated!
 A: Notice first that if $f : U \subset \mathbb{R}^{a} \to \mathbb{R}^{b}$ is a diffeomorphism onto its image, and if $p \in U$, then
$$
\mathrm{d}f(p) : \mathbb{R}^{a} \to \mathbb{R}^{b}
$$
is a linear isomorphism. Indeed, the chain rule shows that
$$
\mathrm{d}f(p)\circ\mathrm{d}f^{-1}(f^{-1}(p)) = \mathrm{id}_{\mathbb{R}^b}
$$
and
$$
\mathrm{d}f^{-1}(f(p))\circ\mathrm{d}f(p) = \mathrm{id}_{\mathbb{R}^a}
$$
Hence, $a = b$, and in you question, you have $n_i = m_i$.
Now, suppose $f : U \subset \mathbb{R}^n \to \mathbb{R}^n$ and $g : V\subset \mathbb{R}^m \to \mathbb{R}^m$ are diffeomorphisms onto their image. Define
$$h : U\times V \to \mathbb{R}^n\times \mathbb{R}^m$$
by $h(p,q) = (f(p),g(q))$. Let us first show that $h$ is continuous. As the vector space topology does not depend on the norm (we are here in finite dimension), let us take the norm $\|x\|_{\mathbb{R}^n} = \sup_{1\leqslant i\leqslant n} |x_i|$. This norm is a good choice because with this norm, we have this fact:
$$
B_{\mathbb{R}^n\times \mathbb{R}^m}((p,q),r) = B_{\mathbb{R}^n}(p,r)\times B_{\mathbb{R}^m}(q,r)
$$
that is, an open ball of radius $r$ in $\mathbb{R}^n\times\mathbb{R}^m$ is the direct product of open balls of radius $r$. As balls generate the topology, to show a function is continuous, it is sufficient to show that the preimage of any ball is open in $U\times V$. Let $r>0$ and $(p,q) \in \mathbb{R}^n\times\mathbb{R}^m$. Then
\begin{align}
h^{-1}\left(B((p,q),r) \right) &= h^{-1} \left(B(p,r)\times B(q,r) \right) \\
&= \left(f^{-1}\left(B(p,r)\right) \times V \right)\cap \left(U\times g^{-1}\left(B(q,r) \right) \right)
\end{align}
Now, as $f$ and $g$ are continuous, $f^{-1}\left(B(p,r) \right)$ is open in $U$, and $g^{-1}\left(B(q,r)\right)$ is open in $V$. It follows that $f^{-1}\left(B(p,r) \right)\times V$ and $U\times g^{-1}\left(B(q,r)\right)$ are open in $U\times V$, and so is their intersection.
This shows that $h$ is continuous.
Now, to show that $h$ is differentiable, let us just fix $(p,q) \in U\times V$ and $(u,v) \in \mathbb{R}^n\times \mathbb{R}^m$ with small norms. Then
\begin{align}
h\left((p,q) + (u,v)\right) &= \left(f(p+u),g(q+v) \right) \\
&= \left(f(p) + \mathrm{d}f(p)u + o(\|u\|),g(q)+\mathrm{d}g(q)v + o(\|v\|) \right) \\
&= (f(p),g(q)) + (\mathrm{d}f(p)u,\mathrm{d}g(q)v) + o(\|(u,v)\|)
\end{align}
and this shows that $h$ is differentiable at $(p,q)$ with $\mathrm{d}h(p,q) = \left(\mathrm{d}f(p),\mathrm{d}h(q) \right)$.
Remark that we could just have shown the differentiability and the continuity would have followed. But I think, in a pedagogical point of view, that showing the continuity by its definition was important.
Now, let $f^{-1}$ and $g^{-1}$ be the inverse of $f$ and $g$ defined on $f(U)$ and $g(V)$ respectively. What we have done before is still valid by replacing $f$ by $f^{-1}$, $g$ by $g^{-1}$, $U$ by $f(U)$ and $V$ by $g(V)$. This shows that if $l = (f^{-1},g^{-1})$, then $l$ is also differentiable and continuous.
Moreover, it is a straightforward calculation that $l\circ h = \mathrm{id}_{U\times V}$ and $h \circ l = \mathrm{id}_{f(U)\times g(V)}$.
To conclude, we have shown that:

*

*$h$ is differentiable map

*$h$ is bijective (onto its image) with invere $l$

*$h^{-1}$ is differentiable

This is exactly saying that $h$ is a diffeomorphism.
A: To see that the Cartesian product of two continuous maps is continuous:
Suppose $f: X \to Y$ and $g: Z \to W$ are continuous, where $X, Y, Z, W$ are (for convenience) metric spaces.  A sequence $(x_n, z_n)$ converges to $(x,z)$ in $X \times Z$ iff $x_n \to x$ in $X$ and $z_n \to z$ in $Y$.  Then $f(x_n) \to f(x)$ in $Y$ and
$g(z_n) \to g(z)$ in $W$, so $(f \times g)(x_n, z_n) = (f(x_n), g(z_n)) \to (f(x),g(z))$ in $Y \times W$.  Thus $f \times g$ is continuous.
As for the inverse, it's easy to see that $(f \times g)^{-1} = f^{-1} \times g^{-1}$.
