Topology of a manifold

Question

A manifold $M$ is a locally euclidian topological space (every point has a neighborhood homeomorphic to an open subset of $\mathbb{R}^n$). We assume, in addition, $M$ Haussdorf and second countable. My question is: are the following properties equivalent?

$M$ is $\sigma$-compact (or countable at infinity).

$M$ has a countable atlas.

$M$ is metrizable and separable.

$M$ is paracompact.

$M$ admits a partition of unity subordinate to any open cover.

Are the following results 'easy' to prove:

1. Any atlas of class $C^k$ ($k\geq1$) is $C^k$-compatible with a $C^\infty$ atlas.

2. Any manifold has a finite atlas (using the theory of dimension?).

Thank you for your help.

Answer 1

About the topology of manifolds we have

paracompactness $\iff$ partitions of unity $\iff$ every component has a countable basis $\iff$ ...

Not hard to prove.

About differentiable structures, it is more convenient to talk about diffeomorphisms. Here are the facts (classic results from the $50$'s) -- (assume paracompact to be safe)

Let $M$ a $C^k$ manifold ($k\ge 1$). Then there exists a $C^{\omega}$ manifold ( real analytic) $\tilde M$ and a $C^k$ diffeomorphism $M \simeq \tilde M$ ( this is equivalent to: there exists a $C^{\omega}$ atlas on $M$ which is $C^k$-compatible with the given $C^k$ atlas.

Let $1\le k \le l$ $M$, $M'$ two $C^l$ manifolds that are isomorphic as $C^k$ manifolds. Then they are isomorphic as $C^l$ manifolds. From the previous statement only the case $k=1$, $l=\omega$ has to be considered.

Whitney showed that every differentiable manifold is a submanifold of some $\mathbb{R}^N$ with image real analytic. This implies the existence of a real analytic structures. John Nash ( of a beautiful mind) proved that every Riemannian manifold can be imbedded in some $\mathbb{R}^n$ with image a real analytic submanifold.

The previous results do not work for $k=0$:

There exists a topological ($C^0$ ) manifold that is not homeomorphic to a ($C^1$) differentiable manifold ( Kervaire).

There exist non-diffeomorphic differetiable manifolds that are homeomorphic (Milnor).

See also exotic $\mathbb{R}^4$.