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In the book General Relativity by Robert Wald, he defines a manifold as follows.

An $n$-dimensional, $C^\infty$, real manifold $M$ is a set together with a collection of subsets $\{O_\alpha\}$ satisfying the following properties:

$(1)$ Each $p\in M$ lies in at least one $O_\alpha$, i.e. $\{O_\alpha\}$ cover $M$.

$(2)$ For each $\alpha$, there is a one-to-one, onto map $\psi_\alpha: O_\alpha\to U_\alpha$, where $U_\alpha$ is an open subset in $\mathbb{R}^n$.

$(3)$ If any two sets $O_\alpha$ and $O_\beta$ overlap, $O_\alpha\bigcap O_\beta\neq \emptyset$, we can consider the map $\psi_\beta\circ\psi_\alpha^{-1}$ which takes points in $\psi_\alpha[O_\alpha\bigcap O_\beta]\subset U_\alpha$ in $\mathbb{R}^n$ to points in $\psi_\beta[O_\alpha\bigcap O_\beta]\subset U_\beta$ in $\mathbb{R}^n$. We require these subsets to be open and this map to be $C^\infty$.

Question Does the manifold, as defined above, make it a topological space?

I don't think so. The set $\{O_\alpha\}$ (defined above) is not a topology unless any arbitrary union and the finite intersection of these subsets $O_\alpha$ belong to this set. According to tot the definition, this need not be satisfied here. Also, there is nothing that says $\emptyset$ and $M$ belong to the set $\{O_\alpha\}$.

I must be missing something because I have seen many people defining a manifold as a topological space.

Sometimes one adds the prefixes "topological" and "differentiable" before the word manifold.

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As soon as you have an atlas ${\cal A}_0$ in Wald's sense (a family of maps $(O_\alpha, \psi_\alpha)$ one-to-one with open sets $U_\alpha$ in $\mathbb{R}^n$ from subsets $O_\alpha$ of $M$ satisfying requirments (1),(2),(3)), $M$ receives a topological structure.

To see it, complete the atlas to a maximal atlas ${\cal A}$ just by adding all possible local charts $\psi: O\to U\subset \mathbb{R}^n$ ($U$ open) that are compatible with all charts of ${\cal A}_0$.

An atlas ${\cal A}$ is maximal when, given a bijective map $\psi: O \to U \subset \mathbb{R}^n$ where $U$ is open, if $(O, \psi)$ is $C^\infty$ compatible with the charts of ${\cal A}$ in the sense of (3), then $(O,\psi) \in {\cal A}$.

It is furthermore easy to prove that there is a unique maximal atlas ${\cal A}$ containing a given atlas ${\cal A}_0$.

Define the ``open sets'' $U \subset M$ as the sets such that for every $p\in U$ there is a local chart $(O,\psi) \in {\cal A}$ with $p \in O\subset U$.

This family of sets (including $\emptyset$ in the family) satisfies all the properties of a topology. Maximality guarantees that the family is closed with respect to finite intersections. That is because the restriction of a chart in ${\cal A}$ to the intersection of a finite number of domains of charts must be an element of ${\cal A}$ as well.

This topology, which is uniquely induced by the initial atlas ${\cal A}_0$, makes $M$ a topological manifold and the initial atlas a topological atlas: the local charts are homeomorphisms from their domains onto their images.

Compatible atlas ${\cal A}_0$, ${\cal A}_0'$ define the same topology as above, which therefore depends only on the common differentiable structure: it is the unique maximal atlas ${\cal A}$ associated to both atlases.

I do not like very much this approach from the mathematical viewpoint because a prefer to keep separated the topological and the differentiable properties. Paracompactness, for instance, arises by further assuming Hausdorff and 2nd countability properties on the topology (also Wald's book assumes these hypotheses later). I prefer to state these hypotheses a priori. However I like it from the physical viewpoint as it reflects the practical use of smooth manifolds in physics.

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  • $\begingroup$ I think this is not quite correct. The charts don't form a basis for the topology, you have to include preimages of the open sets from the charts. So, e.g., you could have a set $U$ that is open but not contained in any chart. See my answer. $\endgroup$ Commented Apr 29, 2021 at 15:42
  • $\begingroup$ I do think this approach is more compatible with physics though. Usually one solves the EFE in a coordinate neighborhood then makes coordinate transformations to "uncover" larger and larger regions of the manifold. See eg. the construction of the maximally extended Schwarzschild solution. In this case the manifold structure and thus the topological structure as well is not fixed from the beginning, but is rather induced by a (non-maximal) system of coordinate neighborhoods. $\endgroup$ Commented Apr 29, 2021 at 16:24
  • $\begingroup$ @Jahan Claes You are right, I forgot to impose maximality for atlases. I made more precise my answer. If the atlas is maximal the "open sets" defined as above are also closed with respect to the finite intersection... $\endgroup$ Commented Apr 29, 2021 at 16:34
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    $\begingroup$ @PLL I think you are right: it is enough to consider the set of all charts compatible with the charts of the atlas: no choice is necessary here! They are automatically compatible to each other and they define the maximal atlas. I remove Zorn's lemma from my answer. $\endgroup$ Commented Apr 30, 2021 at 18:16
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    $\begingroup$ @PLL In fact, my argument was like to kill a fly with a gun (direct translation from Italian). $\endgroup$ Commented Apr 30, 2021 at 18:24
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The $\{O_\alpha\}$ aren't the open sets of the topological space. The open sets are "inherited" from the charts. In other words, a basis of open sets is $\{\psi_\alpha^{-1}(U):U\subseteq U_\alpha\}$, the preimages of open sets in $U_\alpha\subseteq\mathbb{R}^n$.

You can check for yourself that the consistency condition (3) makes this closed under finite intersection. To get all the open sets, you consider arbitrary unions of this basis.

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  • $\begingroup$ I don't understand the point. Charts are the pair $(\psi_\alpha, O_\alpha)$. $\endgroup$
    – SRS
    Commented Apr 29, 2021 at 15:17
  • $\begingroup$ Yes, but the charts are not the open sets. For example, I can define a 1D manifold that is just the open interval $(0,1)$. This requires a single chart, with $O_\alpha$ the whole manifold, and $\psi_\alpha$ the identity operator. $\endgroup$ Commented Apr 29, 2021 at 15:18
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    $\begingroup$ Perhaps it would help OP if you made a bit more explicit that this way of endowing $M$ with a topology is equivalent to the usual definition where we start with $M$ as a topological space and the charts are homeomorphisms. $\endgroup$
    – ACuriousMind
    Commented Apr 29, 2021 at 15:20
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    $\begingroup$ @SRS The point is if you want to define a topology using the charts, you don't say the sets $O_\alpha$ form the topology. The collection $(O_\alpha,\psi_\alpha)$ together allow you to define a topology on your manifold, but it requires more work than just letting $O_\alpha$ be the open sets. $\endgroup$ Commented Apr 29, 2021 at 15:20
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First let us start with the standard definitions. A topological space $M$ is a non-empty set together with a collection of subsets ${\cal T}$ obeying three conditions:

  1. Both $M$ and $\emptyset$ belong to ${\cal T}$;

  2. If ${\cal O}_\lambda\in {\cal T}$ for $\lambda\in \Lambda$, an arbitrary indexing set, then $\bigcup_{\lambda\in \Lambda}{\cal O}_\lambda \in {\cal T}$;

  3. If ${\cal O}_i\in {\cal T}$ for $i\in I$, a finite indexing set, then $\bigcap_{i\in I}{\cal O}_i\in {\cal T}$;

An $n$-dimensional chart on a topological space $(M,{\cal T})$ is a pair $(U,x)$ where $U\in \cal T$ and $x : U\to \tilde{U}\subset \mathbb{R}^n$, with $\tilde{U}$ open in $\mathbb{R}^n$, and with $x$ being a homeomorphism. We say that two charts $(U,x)$ and $(V,y)$ with $U\cap V\neq\emptyset$ are $C^k$-compatible if $x\circ y^{-1}:\tilde{V}\to \tilde{U}$ and $y\circ x^{-1}:\tilde{U}\to \tilde{V}$ are of class $C^k$ according to the usual definition in $\mathbb{R}^n$.

One $n$-dimensional $C^k$-atlas for the topological space $(M,\cal T)$ is a collection ${\cal A}$ of $n$-dimensional charts whose domains cover $M$ and with the property that whenever such domains overlap the charts are $C^k$-compatible.

Finally one $n$-dimensional $C^k$-manifold is a topological space $(M,\cal T)$ together with a maximal $n$-dimensional $C^k$ atlas ${\cal A}$. We may therefore refer to it by the triple $(M,{\cal T},{\cal A})$.

Observe that the topological structure is part of the definition of the manifold. Without the topological structure you cannot judge whether the coordinate maps $x : U\to \tilde {U}$ are homeomorphisms or not.

There is, though, one situation, to which your question pertains. Suppose you have a non-empty set $M$ together with subset $U_i\subset M$ whose union cover $M$. Suppose further you have bijections $x_i : U_i\to \tilde{U}_i\subset\mathbb{R}^n$ and that whenever $U_i\cap U_j\neq\emptyset$ you have that $x_i\circ x_j^{-1}:\tilde{U}_j\to \tilde{U}_i$ and $x_j\circ x_i^{-1}:\tilde{U}_i\to \tilde{U}_j$ are $C^k$-maps. You have everything you need for a $C^k$-manifold, except that you don't yet have a topology on $M$. Could there be some topology which makes these maps homeomorphisms and therefore allows you to get a $C^k$-manifold? The answer is yes, you can construct such a topology and the topology you get is sometimes said to be induced from the charts. That is the situation that Wald has in mind and that is how it connects to the standard definition of a manifold which demands it to be a topological space in the first place.

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You are right that this definition by itself does not make a topology. To get the connection to topology, we need to make an extra definition of what the topology is. What this definition defines is called an atlas, and it says a manifold is a "space with an atlas".

However, the relevant definition to obtain the topological structure is very "natural": The topology on $M$ is the one which makes the maps $\psi_\alpha$ you mention, together with their inverses, into continuous functions. I say this is "natural" because continuous functions are the natural maps (the "morphisms", in category-theoretic terms) between topological spaces, i.e. ones that preserve structure. In effect, you are using the maps that are already part of the definition given to "soak up the topology" from $\mathbb{R}^n$ into the manifold.

But there's no reason we cannot define it differently any more than we could define anything else any way we like - though if we do that, then we will have two inequivalent ideas of what "manifold" means going under the same name.

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    $\begingroup$ This is my favorite way of creating a topology! There's a modified version that I also like. Instead of asking for the unique topology that makes the $\psi_\alpha$ and their inverses continuous, we can ask for the coarsest topology that makes the $\psi_\alpha$ continuous. That gives us the same topology in this case. $\endgroup$
    – Vectornaut
    Commented Apr 30, 2021 at 12:29
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    $\begingroup$ @Vectornaut : Thanks, yep, that would work too. $\endgroup$ Commented May 1, 2021 at 1:42
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Given a manifold, defined as in the question, perhaps the simplest definition of the resulting topology on $M$ is that a set $A\subseteq M$ is open if and only if, for each $\alpha$, $\psi_\alpha(A\cap O_\alpha)$ is open in $U_\alpha$.

Intuitively, the key ingredients leading to this definition are (1) whether a set $A$ is open should be determined locally, on the sets $O_\alpha$, and (2) within any single $O_\alpha$, the topology should match, via $\psi_\alpha$, the ordinary Euclidean topology on $U_\alpha$.

More formally, it's a good idea to check that my definition really produces a topology on $M$ and that the requirements you quoted about the atlas, in particular requirement (3), ensure that each $O_\alpha$ is an open subset in $M$ and that $\psi_\alpha$ is a homeomorphism from $O_\alpha$ (topologized as a subspace of $M$) to $U_\alpha$ (topologized as a subspace of $\mathbb R^n$. It's probably also worthwhile to check that my description of the topology is equivalent to the one in The_Sympathizer's answer and Vectornaut's comment under that answer.

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There's a notion of open sets, a requirement that the manifold be covered by open sets, and a requirement that the intersection of open sets stay open. I'd say the above definition doesn't make sense if the manifold is not a topological space.

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