Dimension of topological manifold and dimension of smooth manifold in Tu Manifolds

Question

Tu Manifolds

In section 5.3, Tu says a "manifold" has dimension $n$ if all of its connected components have dimension n in Definition 5.9 below:

Back in Section 5.1, Tu says in Definition 5.2 that a topological manifold $M$ has dimension $n$ if $M$ is locally Euclidean of dimension $n$.

1. In Definition 5.9, does the "manifold" in "manifold is said to have dimension n" refer to the pair $(M,\mathfrak U)$ of a topological manifold and a maximal atlas instead of just the topological manifold $M$?

- If the answer to 1 is yes:

2. If "connected components" refers to $(M,\mathfrak U)$, then what are "connected components" of something that looks like "$(M,\mathfrak U)$" ?

I think $\mathfrak U$ will turn out to be to M as a topology $\mathscr T$ is to a space $X$, so "connected components" depends on $\mathfrak U$, in differential geometry as in $\mathscr T$ in topology.

3. If "connected components" refers to $M$, then our definition is

A manifold $(M,\mathfrak U)$ has dimension $n$ if the connected components of the topological manifold $M$ are locally Euclidean of dimension $n$.

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4. What is the relationship between $\dim(M)$ and $\dim(M,\mathfrak U)$?

- If the answer to 1 is no:

5. So then this is a proposition instead of a definition

A topological manifold $M$ is locally Euclidean of dimension $n$ if and only if its connected components are locally Euclidean of dimension $n$

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Answer 1

Well, a manifold always comes with the structure of an atlas, but it is far from being a topology, for example, lets take the interval $[0,1)$ and consider the two atlases $$[0,1] \xrightarrow{\iota} \mathbb{R} \quad \textrm{ and } \quad [0,1) \xrightarrow{\textrm{arctan}} \mathbb{R} $$ where $\iota$ is just the canonical inclusion. Then both of those make $[0,1)$ into a differentiable manifold, although they look "fairly" different (the second one makes it look like $\mathbb{R}^+$). So yes, whenever someone says: a manifold $M$, they actually mean $(M',U)$, where $M'$ is a topological space $(M''.T)$ hence no: connectedness does not depend on the atlas! since this is encoded in the topology, that is provided with $M$. Hence since $M$ always means $(M,U)$ the relationship between both dimensions is literally: they are the same, just by definition.

Now you may also realize that, since your charts are homeomorphisms to $\mathbb{R}^n$ and have to be compatible with intersections, one can see that the all charts on the same connected component have the same dimenion. or even better: the dimension at a point defines a continuous map $M \to \mathbb{N}$ and hence they have to agree on connected components.

I hope I answered and understood your problems correctly, if not, please tell me so!