Constructing Smooth Manifolds

Question

I’ve been trying to teach myself differential geometry using Will Merry’s notes (found here) and am struggling to prove Proposition 1.17 on page 7. Here’s the statement…

Let $M$ be a set. Suppose we are given a collection $\{U_a\ | \ a\in A\}$ of subsets of $M$ together with bijections $x_a: U_a \to \mathcal{O}_a$ where $\mathcal{O}_a$ is an open subset of $\mathbb{R}^m$. Assume in addition that:

(I) For any $a,b \in A$, $x_a(U_a \cap U_b)$ is open in $\mathbb{R}^m$

(II) If $U_a \cap U_b \neq \varnothing$, the map $x_b \circ x^{-1}_a: x_a(U_a \cap U_b) \to x_b(U_a \cap U_b)$ is a diffeomorphism

(III) Countably many of the $U_a$ cover $M$

(IV) If $p\neq q$ are points in $m$ then either $\exists a \in A$ so that $p,q \in U_a$ or $\exists a,b \in A$ so that $p \in U_a$, $q \in U_b$, and $U_a \cap U_b=\varnothing$.

Then $M$ has a smooth manifold structure for which the collection $\{x_a : U_a \to \mathcal{O}_a\}$ is an atlas.

For context, Merry defines (1) a topological manifold of dimension $m$ to be a separable and metrizable topological space which is locally homeomorphic to $\mathbb{R^m}$ and (2) a smooth manifold of dimension $m$ to be a topological manifold of dimension $m$ equipped with an atlas which satisfies (II).

Thus I really only need to show (1), i.e. that there is a well-defined separable and metrizable topology on $M$ which makes the $x_a$’s homeomorphisms.

My immediate intuition is to consider the topology generated by the sets: $$\bigcup_{a \in A} \{x_a^{-1}(O) \ | \ O \text{ is an open set of } \mathcal{O}_a\}$$

While this topology clearly makes each $x_a$ continuous, is it possible to further show that the $x_a$’s are actually homeomorphisms? Furthermore, how do I demonstrate the topology is separable and metrizable?

Any help is appreciated, thanks in advance :)

Answer 1

You correctly state that you only need to show that there is a well-defined separable and metrizable topology $\mathscr T$ on $M$ which makes the $x_a$ homeomorphisms.

The idea is that each $x_a$ should be a homeomorphism between an open subset of $M$ and $\mathcal{O}_a$. Thus you correctly argue that as a subbasis of this topology $\mathscr T$ we can take $$\mathscr S =\bigcup_{a \in A} \{x_a^{-1}(O) \ | \ O \text{ is an open set of } \mathcal{O}_a \} = \bigcup_{a \in A} \mathscr S_a$$ with $$\mathscr S_a = \{x_a^{-1}(O) \ | \ O \text{ is an open set of } \mathcal{O}_a \} .$$ Clearly $\mathscr S_a$ is the unique topology on the set $U_a$ making the $x_a : (U_a, \mathscr S_a) \to \mathcal O_a$ homeomorphisms. Moreover, the topology $\mathscr T$ generated by $\mathscr S$ is the coarsest topology making the $x_a : (U_a,\mathscr T_a) \to \mathcal O_a$ continuous, where $\mathscr T_a$ is the subspace induced by $\mathscr T$ on $U_a$ (which is finer than $\mathscr S_a$). It remains to show that $\mathscr T_a = \mathscr S_a$.

1. If $V_a \in \mathscr S_a, V_b \in \mathscr S_b$, then $V_a \cap V_b \in \mathscr S_a \subset \mathscr S$. In particular, $\mathscr S$ is a basis for $\mathscr T$.

Write $V_a = x_a^{-1}(O_a), V_b = x_b^{-1}(O_b) $ with an open $O_a \subset \mathcal O_a, O_b \subset \mathcal O_b$. The bijection $x_b \circ x_a^{-1} : x_a(U_a \cap U_b) \to x_b(U_a \cap U_b)$ is a homeomorphism between open subsets of $\mathbb R^n$. The set $O_b \cap x_b(U_a \cap U_b)$ is open in $x_b(U_a \cap U_b)$, thus $(x_a \circ x_b^{-1})(O_b \cap x_b(U_a \cap U_b)) $ is open in $x_a(U_a \cap U_b)$ and therefore open in $\mathcal O_a$. We conclude that $O_a^* = O_a \cap (x_a \circ x_b^{-1})(O_b \cap x_b(U_a \cap U_b))$ is open in $\mathcal O_a$. But we have $$x_a^{-1}(O^*_a) = x_a^{-1}(O_a) \cap x_b^{-1}(O_b \cap x_b(U_a \cap U_b)) = x_a^{-1}(O_a) \cap x_b^{-1}(O_b) \cap (U_a \cap U_b) \\= x_a^{-1}(O_a) \cap U_a \cap x_b^{-1}(O_b) \cap U_b = x_a^{-1}(O_a) \cap x_b^{-1}(O_b) = V_a \cap V_b .$$ Thus $V_a \cap V_b \in \mathscr S_a$.

2. $\mathscr T_a = \mathscr S_a$.

Trivially $\mathscr S_a \subset \mathscr T_a$. Now consider $V \in \mathscr T_a$. By 1. we can write $V = \bigcup_{j \in J} V_j$ with $V_j \in \mathscr S_{b_j}$. Since $V \subset U_a \in \mathscr S_a$ and $\mathscr S_a$ is a topology, 1. implies $V = \bigcup_{j \in J} V_j \cap U_a \in \mathscr S_a$.

3. $M$ is second countable (i.e. has a countable base).

This follows from the fact that countably many $U_a$ cover $M$ and each $U_a$ has a countable base (because it is homeomorphic to $\mathcal O_a$).

4. $M$ is separable.

It is a well-known theorem of general topology that second countable spaces are separable.

5. $M$ is Hausdorff.

This follows immediately from (IV).

Now there is a problem: The author claims that the proof of Proposition 1.17 is essentially trivial. As Thorgott writes in a comment, the standard definition of a topological manifold requires that is a locally Euclidean Hausdorff second countable space. This was proved above, and this was easy.

The author decided to require that a topological manifold is a locally Euclidean separable metrizable space, and here we have a gap: We must prove that $M$ is metrizable. This can be done by Urysohn's metrization theorem. It says that a regular second countable space is metrizable. This is a non-trival result.

Therefore it remains to prove that $M$ is regular. This is easy. We have to show that for each $ \in M$ and each open neigborhood $U$ of $p$ in $M$ there exists a closed neigborhood $V$ of $p$ in $M$ such that $V \subset U$. W.lo.g. we may assume that $U$ is contained in some $U_a$. The set $U_a$ is locally compact because it is homeomorphic to $\mathcal O_a$. Thus there exist a compact neigborhood $V$ of $p$ in $U_a$ such that $V \subset U$. Since $M$ is Hausdorff, $V$ is a closed neigborhood $V$ of $p$ in $M$.