# Why the product of two manifolds is paracompact?

Some authors define a manifold as a paracompact Hausdorff space that is locally Euclidean. Also it is said that a product of two manifolds is a manifold. However, we know that product of a two paracompact spaces is not necessarily paracompact. So how can we be sure that a product of two manifolds is also paracompact and thus is also a manifold? Is it somehow related to the second-countability property that is usually defined along paracompactness?

## 2 Answers

By the Smirnov metrization theorem, a paracompact Hausdorff space that is locally metrizable is metrizable. Therefore every manifold is metrizable, and hence so is the product of two manifolds. In particular the product is paracompact.

• IIRC this is called the Smirnov metrication theorem – Henno Brandsma Apr 8 '17 at 18:41
• Thank you so much! If I understood it right, it is something like this: Locally Euclidean -> Locally Metrizable -> Metrizable (Hausdorff & paracompact) -> Product of two metrizable spaces is metrizable -> Every metrizable space is paracompact -> Product is paracompact. – Sergey Dylda Apr 8 '17 at 18:44
• @Sergey Yes that's right. – Matt Samuel Apr 8 '17 at 18:45

A connected manifold is paracompact iff it is second-countable, so a general paracompact manifold is just a (possibly uncountable) disjoint union of second-countable manifolds. So, if you take a product of two such spaces, you get a disjoint union of products of two second-countable manifolds. A product of two second-countable spaces is second-countable, so each connected component of the product is second-countable and thus paracompact. So, the product is a disjoint union of paracompact spaces and thus itself paracompact.