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

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:

1. 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.

1. 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 $$n$$.

?

$$\$$

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

# - If the answer to 1 is no:

1. 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 $$n$$

?

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.