In the definition of smooth manifolds, complex manifolds, and similar constructions, one starts by defining a property on neighborhoods in the space, specifying how they relate on overlapping neighborhoods. An atlas is a set of such neighborhoods that covers the space. Some books (Lee, Warner) define the structure as the maximal atlas. Others define it as the equivalence class of compatible atlases.

I was under the impression that the advantage of using the equivalence class definition instead of the maximal atlas definition was that the proof of the existence of such a maximal atlas requires Zorn's lemma, which some prefer not to use if not absolutely necessary.

But Lee and Warner's books both contain existence proofs for this maximal atlas; they start with any atlas, and then just take the set of all compatible charts. If that argument somehow relies on Zorn's lemma (or some other variant of choice), I can't see how. So what do you say? Is choice required, assumed for convenience but not required, or just not necessary at all?

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    $\begingroup$ I did find this thread math.stackexchange.com/questions/43187/why-maximal-atlas where a Mariano Suárez-Alvarez states without justification that Zorn's lemma is not required. $\endgroup$ – ziggurism Sep 22 '11 at 4:04
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    $\begingroup$ It seems clear to me that Zorn's Lemma is not required. $\endgroup$ – Pierre-Yves Gaillard Sep 22 '11 at 4:25
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    $\begingroup$ You think Mariano's answer does not provide justification? Maybe you should justify that claim (sorry...): I found it quite convincing. The point is that in this case there is not just a maximal element in the poset of atlases, there is a unique maximal atlas containing any given one: as you say, you just take the set of all compatible charts. At what point are we choosing anything? $\endgroup$ – Pete L. Clark Sep 22 '11 at 4:28
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    $\begingroup$ By the way, as far as this site is concerned, one should speak of the Mariano Suárez-Alvarez, not a Mariano Suárez-Alvarez. More generally, although I would have to believe that there other other Mariano Suárez-Alvarez's living on Earth, in the mathematical world to the best of my knowledge there is only one... $\endgroup$ – Pete L. Clark Sep 22 '11 at 4:32
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    $\begingroup$ Despite not having deep knowledge in the question's details, I have two points to add to this short discussion: (1) It is completely fine to assume the axiom of choice nowadays. Just as much as you assume that the real numbers have a power set. The controversy about the axiom of choice is long gone. (2) To develop analysis you need some amount of choice. The larger the cardinality of the spaces you have, the more choice you are going to need to have other things behave as you would have wanted. Zorn's lemma can be replaced by a reduced form to match that amount of choice used to begin with $\endgroup$ – Asaf Karagila Sep 22 '11 at 6:17

Zorn's lemma is not required to prove the existence of a maximal atlas, though it is convenient. For one thing, we don't have to prove that compatibility of atlases is an equivalence relation. On the other hand, the obvious proof using Zorn's lemma requires some extra work to show that there is a unique maximal atlas containing any atlas. So let's do it without Zorn's lemma.

Definition. Two atlases on a manifold are compatible if their union is an atlas.

Lemma. Compatibility of atlases is an equivalence relation.

Proof. It is clear that compatibility is symmetric and reflexive, and it remains to be shown that compatibility is transitive. Let $\mathcal{A}_1, \mathcal{A}_2, \mathcal{A}_3$ be three atlases on a $k$-manifold $M$, and suppose $\mathcal{A}_1 \cup \mathcal{A}_2$ and $\mathcal{A}_2 \cup \mathcal{A}_3$ are atlases. We wish to show $\mathcal{A}_1 \cup \mathcal{A}_3$ is an atlas. So let $\varphi_1 : U_1 \to \mathbb{R}^k$ be a chart in $\mathcal{A}_1$, $\varphi_3 : U_3 \to \mathbb{R}^k$ be a chart in $\mathcal{A}_3$. $\mathcal{A}_2$ is an atlas, so for each point $x$ in $U_1 \cap U_3$ there is a chart $\varphi_2 : U_2 \to \mathbb{R}^k$ in $\mathcal{A}_2$ such that $x \in U_2$; but $\varphi_1$ and $\varphi_2$ are compatible and $\varphi_2$ and $\varphi_3$ are compatible so we see that $\varphi_1$ and $\varphi_3$ are locally compatible at $x$. (Here, ‘compatible’ means that the transition map satisfies the relevant regularity condition. There may well be invocations of the axiom of choice hidden here, but I will assume that there are not.) Moreover, $x$ is arbitrary in $U_1 \cap U_3$ so this shows $\varphi_1$ and $\varphi_3$ are compatible; and $\varphi_1$ and $\varphi_3$ are also arbitrary, so $\mathcal{A}_1 \cup \mathcal{A}_3$ is an atlas.

Lemma. The class of atlases on a manifold is a set.

Proof. The class of atlases is a subclass of the set $$\mathscr{P} \left( \bigcup_{U \in \mathscr{P}(M)} \{ U \to \mathbb{R}^k \} \right)$$ where $\{ U \to \mathbb{R}^k \}$ denotes the set of all functions $U \to \mathbb{R}^k$, so by the axiom of separation, the class of atlases is a set.

Lemma. The union of arbitrarily many pairwise compatible atlases is an atlas.

Proof. Immediate.

Theorem. Every atlas is contained in a unique maximal atlas.

Proof. From the above, it is clear that every atlas $\mathcal{A}$ is contained in some equivalence class of atlases, and this equivalence class is a set of compatible atlases. Let $\overline{\mathcal{A}}$ be the union of all those atlases. Then $\mathcal{A} \subseteq \overline{\mathcal{A}}$, and $\overline{\mathcal{A}}$ is the unique maximal atlas containing $\mathcal{A}$: for if $\mathcal{A} \subseteq \mathcal{A}'$, then $\mathcal{A}$ and $\mathcal{A}'$ are compatible, so $\mathcal{A}' \subseteq \overline{\mathcal{A}}$ by construction.

For the sake of completeness I sketch a proof using Zorn's lemma.

Theorem. Every atlas is contained in a maximal atlas.

Proof. The set of all atlases containing $\mathcal{A}$ partially ordered by inclusion is a chain-complete poset: indeed, it is clear that if we have a chain $\{ \mathcal{A}_\alpha \}$, then $\bigcup_\alpha \mathcal{A}_\alpha$ is also an atlas. Thus, the hypotheses of Zorn's lemma are satisfied and there is some maximal atlas containing $\mathcal{A}$.

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  • $\begingroup$ Hi Zhen Lin. Thank you for that very detailed response. So AC is convenient, but not required. When Lee and Warner wave their hands and say "just take the set containing all compatible charts", they don't mention equivalence relations. Actually, Lee says something along the lines of "we could use an equivalence relation, but let's not". How much are they allowed to skip? Do you think it's rigorous to skip the equivalence relation stuff, and also skip Zorn's lemma? I guess the tradeoff is that you don't get uniqueness, is that right? $\endgroup$ – ziggurism Sep 23 '11 at 0:15
  • $\begingroup$ @ziggurism: I don't think it's entirely legitimate to skip the transitivity proof. There's no a priori reason why the set of all atlases compatible with $\mathcal{A}$ should be a set of pairwise compatible atlases; but once we establish that they are all pairwise compatible, we can just take the union of them. Perhaps what's more important is establishing that there is in fact a set of atlases to take the union of: the axioms of ZF do not permit us to take a class-indexed union. $\endgroup$ – Zhen Lin Sep 23 '11 at 0:44
  • $\begingroup$ @ZhenLin what is axiom of seperation and how it is used here? teach me please. $\endgroup$ – Marso Aug 22 '12 at 0:03
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    $\begingroup$ It is the first time I see in writing a proof that compatibility of atlases is transitive (while compatibility of charts is not transitive). Congratulations for your sense of rigor, Zhen! $\endgroup$ – Georges Elencwajg Aug 22 '12 at 9:28
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    $\begingroup$ @Flute The axiom of separation is an axiom (schema) of Zermelo–Fraenkel set theory, and you've probably been using it all along without realising it! $\endgroup$ – Zhen Lin Aug 22 '12 at 11:23

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