The point of this problem is to help the reader understand the difference between two distinct concepts. Suppose $M$ is a topological manifold, and $\mathscr A_1$ and $\mathscr A_2$ are two different smooth structures on $M$ (i.e., maximal smoothly compatible atlases). Then we can ask two questions about $\scr A_1$ and $\scr A_2$:
- What does it mean for $\scr A_1$ and $\scr A_2$ to be the same smooth structure on $M$?
- What does it mean for the smooth manifolds $(M,\scr A_1)$ and $(M,\scr A_2)$ to be diffeomorphic to each other?
In question 1, we are given two different atlases on $M$, and the question is whether each chart of $\scr A_1$ is smoothly compatible with each chart in $\scr A_2$ and vice versa. The problem you quoted (Problem 1-6) asks you to construct uncountably many smooth structures on a given manifold that are distinct, in the sense that the charts of one are not smoothly compatible with the charts of another. This is possible even on $\mathbb R$; indeed, as the problem states, it is possible on any positive-dimensional topological manifold as long as it admits at least one smooth structure.
The second question (whether two given smooth structures on $M$ result in smooth manifolds that are diffeomorphic to each other) is a completely different question. The result that @levap refers to about $\mathbb R$ (Problem 15-13 in my book) says that if $\scr A_1$ and $\scr A_2$ are two smooth structures on $\mathbb R$, then there is a map $F\colon \mathbb R\to\mathbb R$ that is a diffeomorphism from $(\mathbb R,\scr A_1)$ to $(\mathbb R,\scr A_2)$. Another way of saying this is that given a smooth chart $(U,\phi)\in\scr A_2$, the chart $(F^{-1}(U),\phi\circ F)$ will be a chart for $\mathbb R$ that is compatible with all the charts in $\mathscr A_1$ (and thus, by maximality, is already in $\mathscr A_1$). It doesn't say that every chart in $\mathscr A_2$ is already smoothly compatible with those in $\mathscr A_1$. And it does not contradict the fact that there are many distinct smooth structures on $\mathbb R$; it just says that any two of them are related to each other by such a map.
For Question 1, you are right. For instance, you can just take the set of all charts on $(M,T)$ and they will be an atlas.
For Questions 2 and 3, as you have defined a topological manifold, a topological manifold is just a topological space which satisfies certain properties. So, an atlas doesn't actually have anything to do with what a topological manifold is (an atlas just happens to exist on any topological manifold). Two manifolds are equal iff they are equal as topological spaces.
That said, no one actually cares about equality of manifolds. What people actually care about is whether two manifolds are homeomorphic (or more specifically, whether specific maps between them are homeomorphisms). In other words, the "naive equivalence" you are asking about is not important for any applications. As a result, it's perfectly fine to use a definition as you propose in Question 3, where an atlas is part of what a manifold is. This will change what equality of manifolds means (i.e., "naive equivalence") but will not change the notion of equivalence that actually matters, which is homeomorphism.
In the language of category theory, you can define a category $Man$ whose objects are topological manifolds (according to your original definition) and whose maps are continuous maps. You can also define a category $Man'$ whose objects are topological manifolds together with an atlas and whose maps are continuous maps. There is a forgetful functor $F:Man'\to Man$ which forgets the atlas. This functor is not an isomorphism of categories, but it is an equivalence of categories, which is good enough for everything people ever want to do with manifolds.
As a final remark, atlases are pretty irrelevant to the study of topological manifolds. The reason atlases are important is to define smooth manifolds, which impose some additional conditions on what kind of atlases are allowed. A smoooth manifold cannot be defined as just a topological space, but instead must be defined as a topological space together with an atlas satisfying certain assumptions (or a topological space together with some other additional structure equivalent to an atlas).
For smooth manifolds, although an atlas must be included in the definition, there is still an issue similar to your Questions 2 and 3. Namely, multiple different atlases can give "the same" smooth manifold, in the sense that the identity map is a diffeomorphism. This means that if you define a smooth manifold as a triple $(M,T,A)$ where $(M,T)$ is a topological space and $A$ is a smooth atlas on $(M,T)$, then the "naive equivalence" is not the equivalence you actually care about, similar to if you used the definition for topological manifolds you proposed in Question 3.
To avoid this, many authors instead define a smooth manifold as a triple $(M,T,A)$ where $(M,T)$ is a topological space and $A$ is a maximal smooth atlas on $(M,T)$ (or alternatively, $A$ is an equivalence class of smooth atlases on $(M,T)$). This makes the choice of $A$ unique, in the sense that if $(M,T,A)$ and $(M,T,A')$ are smooth manifolds such that the identity map $M\to M$ is a diffeomorphism between them, then $A=A'$. As with topological manifolds, though, it doesn't really matter whether you use this definition or the previous one, since all that changes is what it means for two smooth manifolds to literally be equal and that's not what we actually care about.
Best Answer
I don’t know how much Carroll has introduced at this stage, but a smooth manifold consists at the very minimum of the following information: an integer $n\geq 0$, a set $M$ together with a choice $\mathcal{A}$ called an atlas. An atlas is an object whose elements are pairs, $(U,\phi)$, called charts. These charts are then required to satisfy a few conditions:
Ok I’m glossing over the fact that after one has the definition of $\mathcal{A}$, a smooth structure on $M$ is actually an equivalence class of atlases (the equivalence relation being that the union must be an atlas still) or what amounts to the same thing, one considers a maximal atlas.
Anyway, the point of this question is for you to think about if it is possible to take the set $M=\Bbb{R}^n$ (forget everything else you know about it temporarily; i.e forget grid lines, forget Euclidean distances, forget all of that. Just think of this as a big box of tuples) and construct an atlas $\mathcal{A}$ such that with this atlas, $(M,[\mathcal{A}])$ is a smooth manifold of dimension $m$ (with $m$ possibly different from $n$), and furthermore whether it is possible to make it so that $(M,[\mathcal{A}])$ is diffeomorphic to the usual smooth manifold $(\Bbb{R}^m,[\mathcal{A}_{\text{standard}}])$.
I’ll spoil the answer for you, but let you figure out the details (I disagree that you cannot answer this rigorously at this stage (even if you don’t know what a topological space in general is, as long as you know what an atlas is and what the usual topology on $\Bbb{R}^n$, in terms of open balls is)). The answer is yes, such an atlas $\mathcal{A}$ does exist. In fact, I can even take the Cantor set (or even worse, a fat Cantor set $F\subset\Bbb{R}$) and I can equip it with an atlas such that the result is diffeomorphic to $\Bbb{R}^{2023}$ with its usual smooth structure.
So, the purpose of this question, as already mentioned in the text, is “to provoke you to think deeply about what a manifold is”, and in particular, to get you to keep your instincts at bay and proceed only with the definitions in an almost robotic manner “smooth manifold = set + maximal smooth atlas” and to make you understand the role of each part of the definition. In this process you’ll learn the key lesson that a manifold is not just a set of points!