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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?

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  • $\begingroup$ $\mathcal{O}$ 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 $\mathcal{O}$? $\endgroup$
    – Paul Frost
    Sep 28, 2018 at 16:04
  • $\begingroup$ Yes, at least if they are equivalent. $\endgroup$
    – user370967
    Sep 28, 2018 at 16:53
  • $\begingroup$ 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 $b_*(\mathcal{A})$! $\endgroup$
    – Paul Frost
    Sep 28, 2018 at 17:02
  • $\begingroup$ And my proof is correct then? $\endgroup$
    – user370967
    Sep 28, 2018 at 17:09
  • $\begingroup$ What you have proved is correct. See Seub's answer. $\endgroup$
    – Paul Frost
    Sep 28, 2018 at 17:10

1 Answer 1

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(1) Yes, it's entirely correct, good work.

(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.

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  • $\begingroup$ Thank you. I will ask my professor why his/her proof proves that $\mathcal{O}$ doesn't depend on the chosen atlas. $\endgroup$
    – user370967
    Sep 28, 2018 at 17:11

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