I was trying to prove the following proposition, which is left as exercise in Tu's Introduction To Manifolds.

Proposition 5.18 (An atlas for a product manifold). If $\left\{ (U_{\alpha}, \phi_{\alpha}) \right\}$ and $\left\{ (V_i, \psi_i) \right\}$ are $C^{\infty}$ atlases for the manifolds $M$ and $N$ of dimension $m$ and $n$, respectively, then the collection $$ \left\{(U_\alpha \times V_i, \phi_{\alpha} \times \psi_i : U_{\alpha} \times V_i \to \mathbb{R}^m \times \mathbb{R}^n \right\} $$ of charts is a $C^{\infty}$ atlas $M \times N$. Therefore, $M\times N$ is a $C^{\infty}$ manifold of dimension $m + n$.

I provide also the definitions I believe I should use.

Definition 5.1. : A topological space $M$ is locally Euclidean of dimension $n$ if every point $p$ in $M$ has a neighborhood $U$ such that there's is a homeomorphism $\phi$ from $U$ onto an open subset of $\mathbb{R}^n$. We call the pair $(U,\phi : U \to \mathbb{R}^n)$ a chart, $U$ a coordinate neighborhhod or a coordinate open set, and $\phi$ a coordinate map or coordinate system on $U$. We say that a chart $(U,\phi)$ is centered at $p \in U$ if $\phi(p) = 0$.

Definition 5.2 A topological manifold is Hausdorff, second countable, locally Euclidean space. It is said to be of dimension $n$ if it is locally Euclidean of dimension $n$.

I think the proposition is straightforward to prove, because the maps $\phi_\alpha$, and $\psi_i$ are homeomorphism then the product $\phi_\alpha \times \psi_i$ is also an homeomorphism on the product topology induced by the family $U_\alpha \times V_i$ (which is the generic open set). So the map $\phi_\alpha \times \psi_i$ maps the open $U_\alpha \times V_i$ into the product of two open sets $\mathcal{O}_{\mathbb{R}^m} \times \mathcal{O}_{\mathbb{R}^n}$ which is an open set by definition of product topology.

Am I missing anything, or is it simple as that?


It is that easy. The product $M\times N$ is Hausdorff and second countable for any two Hausdorff and second countable spaces $M, N$. And the product of two charts is a chart.

  • $\begingroup$ So my argument was correct, right? $\endgroup$ Oct 9 '19 at 3:22
  • 1
    $\begingroup$ Yes, of course! Trust yourself! $\endgroup$
    – Paul Frost
    Oct 9 '19 at 8:20

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