Topology of a manifold

Often a manifold is defined as a topological space and the domains of the charts are required to be open sets and the co-ordinates homeomorphisms. This is really superfluous as the topology is forced once we have chosen the atlas. Given a manifold $M$ we define a subset $W \subset M$ to be open if for every $x \in W$ there is a chart with domain $U$ such that $x \in U \subset W$. We need to show that such a definition of open sets defines a topology on $M$. The only problem is showing that the intersection of two open sets is open. This follows from the following Lemma whose proof we leave as an exercise.

Lemma 2.1   Let $(U, \psi)$ be a co-ordinate chart on a manifold $M$ and let $W \subset U$ be such that $\psi(W) \subset \psi(U)$ is open. Then $(W, \psi_{\vert W})$ is a co-ordinate chart.

We also leave as an exercise showing that with this topology if $(U, \psi)$ is a co-ordinate chart then $\psi\colon U \to \psi(U)$ is a homeomorphism.

We will in general require a manifold to be Hausdorff and paracompact in the topology.

Now that we have defined the topology of a manifold we can discuss its dimension. Each co-ordinate function has as range some $\mathbb{R}^d$. From the definition of compatibility it is clear that $d$ is constant on the connected components of $M$. We shall go further and assume that our manifolds are such that this number $d$ is constant on all of $M$. We call it the dimension of the manifold.