Canonic topology on differentiable structure characterisation.

Please bear with me. The question might look huge but I just included the relevant definitions for making the post self-contained:

My definition of atlas:

Let $$M$$ be a set. We call a set $$\mathcal{A} = \{(U_i, \phi_i)\mid i \in I\}$$ of local charts an atlas of dimension $$m$$ on $$M$$ if the following conditions are satisfied:

(1) $$U_i \subseteq M$$

(2) $$\phi_i: U_i \to \mathbb{R}^m$$ is injective.

(3) $$\phi(U_i)$$ is open.

(4) $$\bigcup_{i \in I} U_i = M$$

(5) If $$i,j \in I, U_i \cap U_j \ne \emptyset$$, then $$\phi_i(U_i \cap U_j)$$ is open.

(6) If $$i,j \in I, U_i \cap U_j \ne \emptyset$$. Then $$\phi_j \circ \phi_i^{-1}: \phi_i(U_i \cap U_j) \to \phi_j(U_i \cap U_j)$$ is smooth (of class $$C^\infty$$).

Definition of canonic topology:

Given an atlas $$\mathcal{A}$$ of $$M$$, we define the canonic topology on $$(M, \mathcal{A})$$ as the set of all unions of domains of local charts equivalent with $$\mathcal{A}$$. This topology does not change if we replace $$\mathcal{A}$$ by an equivalent atlas.

I want to prove:

Let $$M$$ be a set with atlas $$\mathcal{A} = \{(U_a, \phi_a)\}_{a \in A}$$. Then the canonic topology on $$(M, \mathcal{A})$$ is given by $$\mathcal{O}:= \{V \subseteq M \mid \forall a \in A: \phi_a(V \cap U_a) \mathrm{\ open}\}$$

My attempt:

Write $$\mathcal{T}$$ for the canonic topology. We prove that $$\mathcal{T} \subseteq \mathcal{O}$$.

Let $$V \in \mathcal{T}$$. Then $$V = \bigcup_{i \in I} O_i$$ for local charts $$(O_i, \psi_i)$$ equivalent with $$\mathcal{A}$$. It is easy to see that $$\mathcal{O}$$ is closed under unions, so it suffices to show that $$O_i \in \mathcal{O}$$.

For this, we need to show that $$\phi_a(O_i \cap U_a)$$ is open for every $$a \in A, i \in I$$.

Fix $$i \in I$$. By compability of the chart $$(O_i, \psi_i)$$, we have that $$\mathcal{A} \cup \{(O_i, \psi_i)\}$$ is an atlas, and for every $$a \in A$$, $$\phi_a(O_i \cap U_a)$$ is open, by definition of atlas (more specifically, see (5) in my definition)

For the other inclusion, let $$V \in \mathcal{O}$$. Then $$\phi_a(V \cap U_a)$$ is open for every $$a \in A$$ and

$$V= V \cap M = V \cap (\bigcup_{a \in A} U_a) = \bigcup_{a \in A} (V \cap U_a)$$

so it suffices to show that there are local charts with domain $$V \cap U_a$$ compatible with $$\mathcal{A}$$.

For this, consider the chart $$(U_a \cap V, \phi_a\vert_{U_a \cap V})$$. Because $$V \in \mathcal{O}$$, we have that $$\phi_a(U_a \cap T)$$ is open, so this is a local chart of $$M$$.

It remains to check that it is compatible with $$\mathcal{A}$$. But

$$\phi_a(U_a \cap V \cap U_b) = \phi_a(U_a \cap V) \cap \phi_a( U_a \cap U_b)$$ is open, as intersection of open sets. Similarly, $$\phi_b(U_a \cap V \cap U_b)$$ is open and it is also straightforward to check that the transitions between the local charts are $$C^\infty$$. Hence, $$(U_a \cap V, \phi_a\vert_{U_a \cap V})$$ is compatible with $$\mathcal{A}$$ for all $$a \in A$$ and we are done.

Questions:

(1) Is this proof correct?

(2) I essentially gave the proof my textbook provided, but there they proved that $$\mathcal{O}$$ is a topology and that $$\mathcal{O}$$ does not depend on the chosen atlas. Why is this necessary to prove this?

• $\mathcal{O}$ does not depend on the chosen atlas: Does this mean that any two atlases $\mathcal{A}, \mathcal{A}'$ on the set $M$ should induce the same $\mathcal{O}$? Sep 28, 2018 at 16:04
• Yes, at least if they are equivalent.
– user370967
Sep 28, 2018 at 16:53
• Okay, then it is clear. For non-equivalent atlases you will (in general) get different topologies. Take any bijection $b$ on $M$ to define an atlas $b_*(\mathcal{A})$! Sep 28, 2018 at 17:02
• And my proof is correct then?
– user370967
Sep 28, 2018 at 17:09
• What you have proved is correct. See Seub's answer. Sep 28, 2018 at 17:10

(2) It's never necessary to do anything in life. But for example, proving first that $$\mathcal{O} = \mathcal{T}$$ saves the trouble of proving that $$\mathcal{T}$$ itself is a topology, if you show that $$\mathcal{O}$$ is a topology, which seems a bit easier. On the other hand, it's easier to argue that $$\mathcal{T}$$ does not depend on the atlas than $$\mathcal{O}$$. So it's convenient to have both characterizations of that topology, I guess.
• Thank you. I will ask my professor why his/her proof proves that $\mathcal{O}$ doesn't depend on the chosen atlas.