# Is Zorn's lemma required to prove the existence of a maximal atlas on a manifold?

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?

• 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. Sep 22, 2011 at 4:04
• It seems clear to me that Zorn's Lemma is not required. Sep 22, 2011 at 4:25
• 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? Sep 22, 2011 at 4:28
• 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... Sep 22, 2011 at 4:32
• 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 Sep 22, 2011 at 6:17

## 1 Answer

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 $\mathcal{A}$.

• 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? Sep 23, 2011 at 0:15
• 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! Aug 22, 2012 at 9:28
• @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! Aug 22, 2012 at 11:23
• @Flute It isn't very important. Roughly speaking, if you can describe a subclass of a set by writing down a logical formula that its members, then that subclass is also a set. Aug 22, 2012 at 15:55
• @ZhenLin : You can skip the transitivity proof because you jumped to the set of all charts compatible with the original atlas. No other atlases are considered, so no equivalence relation; and it's obviously a set (not a proper class) since it's a set of partial functions from one fixed set (the set of points) to another ($\mathbb R^n$), such as people form all the time. This doesn't mean that you really skip the hard part of the proof; you have to prove that this set of charts is an atlas, which takes about the same amount of work. But you didn't have to use an equivalence relation. Feb 18 at 18:37